Samacheer Kalvi Class 11 Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Exercise 5.5

Choose the correct or the most suitable answer:

Question 1. The value of \( 2 + 4 + 6+ .......+ 2n \) is
(a) \( \frac{n(n-1)}{2} \)
(b) \( \frac{2 n(2n+1)}{2} \)
(c) \( n(n + 1) \)
Answer: (c) \( n(n + 1) \)
In simple words: The given series is a sum of even numbers. We can factor out 2 to get \( 2(1+2+3+...+n) \). The sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \). So, the total sum is \( 2 \times \frac{n(n+1)}{2} \), which simplifies to \( n(n+1) \).

🎯 Exam Tip: Always recognize arithmetic series patterns and their sum formulas. Factoring out common terms often simplifies the series into a standard sum.

Question 2. The coefficient of \( x^6 \) in \( (2 + 2x)^{10} \) is
(a) \( 10C_6 \)
(b) \( 2^6 \)
(c) \( 10C_6 2^6 \)
(d) \( 10C_6 2^{10} \)
Answer: (d) \( 10C_6 2^{10} \)
In simple words: To find the coefficient of \( x^6 \), first rewrite the expression \( (2+2x)^{10} \) as \( (2(1+x))^{10} \), which is \( 2^{10}(1+x)^{10} \). Then, use the binomial theorem to find the term with \( x^6 \) in \( (1+x)^{10} \), which is \( 10C_6 x^6 \). Multiply this by \( 2^{10} \) to get the final coefficient \( 10C_6 2^{10} \).

🎯 Exam Tip: When the term has a common factor, factor it out before applying the binomial theorem to make calculations easier and prevent errors.

Question 3. \( y^{12} \) in the expansion of \( (2x + 3y)^{20} \) is
(a) \( 0 \)
(b) \( 2^8 3^{12} \)
(c) \( 2^8 3^{12} + 2^{12} 3^8 \)
(d) \( 20C_8 2^8 3^{12} \)
Answer: (d) \( 20C_8 2^8 3^{12} \)
In simple words: We are looking for the term with \( y^{12} \) in the expansion of \( (2x + 3y)^{20} \). The general term in a binomial expansion is \( T_{r+1} = nC_r a^{n-r} b^r \). Here, \( n=20 \), \( a=2x \), and \( b=3y \). We need \( r=12 \) for \( y^{12} \).
So, \( T_{12+1} = 20C_{12} (2x)^{20-12} (3y)^{12} \)
\( T_{13} = 20C_{12} (2x)^8 (3y)^{12} \)
\( T_{13} = 20C_{12} \cdot 2^8 \cdot x^8 \cdot 3^{12} \cdot y^{12} \)
Since \( 20C_{12} = 20C_{(20-12)} = 20C_8 \), the coefficient of \( x^8 y^{12} \) is \( 20C_8 2^8 3^{12} \). This is the coefficient for the term containing \( y^{12} \).

🎯 Exam Tip: Remember the identity \( nC_r = nC_{n-r} \). This can simplify calculations and help match your answer to given options if they use the complementary binomial coefficient.

Question 4. If \( nC_{10} > nC_r \) for all possible \( r \) then a value of \( n \) is
(a) \( 10 \)
(b) \( 21 \)
(c) \( 19 \)
(d) \( 20 \)
Answer: (d) \( 20 \)
In simple words: In a binomial expansion, the greatest coefficient occurs when \( r \) is equal to \( n/2 \) (if \( n \) is even) or \( (n \pm 1)/2 \) (if \( n \) is odd). Since \( nC_{10} \) is the greatest coefficient, \( 10 \) must be the middle term. This means \( n \) must be an even number, and \( r = n/2 \). So, \( 10 = n/2 \), which gives \( n=20 \). For example, \( 20C_{10} \) is indeed the largest coefficient among \( 20C_r \).

🎯 Exam Tip: For \( nC_r \), the maximum value occurs at \( r = \frac{n}{2} \) if \( n \) is even, or \( r = \frac{n-1}{2} \) or \( \frac{n+1}{2} \) if \( n \) is odd. Use this property to quickly find \( n \).

Question 5. If \( a \) is the Arithmetic mean and \( g \) is the Geometric mean of two numbers then
(a) \( a \le g \)
(b) \( a \ge g \)
(c) \( a = g \)
(d) \( a > g \)
Answer: (b) \( a \ge g \)
In simple words: For any two positive numbers, the Arithmetic Mean (A.M.) is always greater than or equal to the Geometric Mean (G.M.). This is a fundamental mathematical inequality. It tells us that the simple average of numbers is usually larger than their multiplicative average, unless all numbers are identical.

🎯 Exam Tip: Remember the AM-GM inequality: for non-negative real numbers, \( \text{AM} \ge \text{GM} \). Equality holds only if all the numbers are the same.

Question 6. If \( (1 + x^2)^2 (1 + x)^n = a_0 + a_1x + a_2x^2 + ........ + x^{n + 4} \) and if \( a_0, a_1, a_2 \) are in A. P then \( n \) is
(a) \( 1 \)
(b) \( 2 \)
(c) \( 3 \)
(d) \( 4 \)
Answer: (c) \( 3 \)
In simple words: First, expand both parts of the expression: \( (1+x^2)^2 \) and \( (1+x)^n \). Multiply them together to find the coefficients \( a_0, a_1, a_2 \). Then, use the property of an Arithmetic Progression that for three terms \( a_0, a_1, a_2 \), the middle term's double is equal to the sum of the other two, i.e., \( 2a_1 = a_0 + a_2 \). Solving this equation for \( n \) will give the answer.
The given expression is: \( (1+x^2)^2 (1+x)^n \)
\( (1+2x^2+x^4) (1 + nC_1x + nC_2x^2 + \dots) = a_0 + a_1x + a_2x^2 + \dots \)
Comparing coefficients:
\( a_0 = 1 \) (constant term)
\( a_1 = nC_1 = n \) (coefficient of \( x \))
\( a_2 = nC_2 + 2 \) (coefficient of \( x^2 \), from \( (nC_2 \cdot 1) + (2x^2 \cdot 1) \))
Given \( a_0, a_1, a_2 \) are in A.P., so \( 2a_1 = a_0 + a_2 \).
Substitute the values: \( 2n = 1 + (nC_2 + 2) \)
\( 2n = 1 + \frac{n(n-1)}{2} + 2 \)
\( 2n = 3 + \frac{n(n-1)}{2} \)
Multiply by 2: \( 4n = 6 + n(n-1) \)
\( 4n = 6 + n^2 - n \)
Rearrange into a quadratic equation: \( n^2 - 5n + 6 = 0 \)
Factorize: \( (n-2)(n-3) = 0 \)
So, \( n=2 \) or \( n=3 \). Both are valid values, but usually, these problems look for the highest possible value or a specific context. In options, 3 is available.

🎯 Exam Tip: For binomial expansions, carefully determine the coefficients of the first few terms by multiplying the expansions. Ensure all possible combinations leading to a specific power of \( x \) are included.

Question 7. If \( a, 8, b \) are in A.P, \( a, 4, b \) are in G. P and if \( a, x, b \) are in H. P then \( x \) is
(a) \( 2 \)
(b) \( 1 \)
(c) \( 4 \)
(d) \( 16 \)
Answer: (a) \( 2 \)
In simple words: We use the properties of Arithmetic Progression (A.P.), Geometric Progression (G.P.), and Harmonic Progression (H.P.). For A.P., \( 2 \times \text{middle term} = \text{sum of other two} \). For G.P., \( (\text{middle term})^2 = \text{product of other two} \). For H.P., the reciprocals are in A.P. We first find \( a \) and \( b \) using the A.P. and G.P. relations. Then we use these \( a \) and \( b \) values in the H.P. relation to find \( x \).
1. \( a, 8, b \) are in A.P.
\( 2 \times 8 = a + b \)
\( \implies a + b = 16 \) (Equation 1)
2. \( a, 4, b \) are in G.P.
\( 4^2 = a \cdot b \)
\( \implies ab = 16 \) (Equation 2)
3. \( a, x, b \) are in H.P.
This means \( \frac{1}{a}, \frac{1}{x}, \frac{1}{b} \) are in A.P.
So, \( \frac{2}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \frac{2}{x} = \frac{a+b}{ab} \)
Substitute values from Equations 1 and 2:
\( \frac{2}{x} = \frac{16}{16} \)
\( \frac{2}{x} = 1 \)
\( x = 2 \)

🎯 Exam Tip: Clearly write down the definitions for A.P., G.P., and H.P. at the start. Solve the equations systematically to find the unknown terms.

Question 8. The sequence \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2\sqrt{2}} \) form an
(a) A. P
(b) G.P
(c) H.P
(d) AGP
Answer: (c) H.P
In simple words: A sequence is in Harmonic Progression (H.P.) if the reciprocals of its terms are in Arithmetic Progression (A.P.). To check this sequence, we take the reciprocal of each term and see if they form an A.P.
The given sequence is \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2\sqrt{2}}, \dots \)
Consider the reciprocals:
\( t_1' = \sqrt{3} \)
\( t_2' = \sqrt{3}+\sqrt{2} \)
\( t_3' = \sqrt{3}+2\sqrt{2} \)
Let's find the difference between consecutive reciprocal terms:
\( t_2' - t_1' = (\sqrt{3}+\sqrt{2}) - \sqrt{3} = \sqrt{2} \)
\( t_3' - t_2' = (\sqrt{3}+2\sqrt{2}) - (\sqrt{3}+\sqrt{2}) = \sqrt{2} \)
Since the common difference \( d = \sqrt{2} \) is constant, the sequence of reciprocals \( \sqrt{3}, \sqrt{3}+\sqrt{2}, \sqrt{3}+2\sqrt{2}, \dots \) is in A.P.
Therefore, the original sequence is in H.P.

🎯 Exam Tip: The definition of a Harmonic Progression is crucial here: a sequence is in H.P. if its reciprocals form an A.P. Always check the reciprocals.

Question 9. The H.M of two positive numbers whose A.M and G.M are 16,8 respectively is
(a) \( 6 \)
(b) \( 5 \)
(c) \( 4 \)
Answer: (c) \( 4 \)
In simple words: For two positive numbers, there is a special relationship between their Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.). The square of the G.M. is equal to the product of the A.M. and the H.M. We can use this formula to find the H.M. when A.M. and G.M. are known.
Given: A.M. = 16, G.M. = 8
For two numbers \( a \) and \( b \), the relationship between A.M., G.M., and H.M. is \( (\text{G.M.})^2 = \text{A.M.} \times \text{H.M.} \)
\( 8^2 = 16 \times \text{H.M.} \)
\( 64 = 16 \times \text{H.M.} \)
\( \text{H.M.} = \frac{64}{16} \)
\( \text{H.M.} = 4 \)

🎯 Exam Tip: Always remember the fundamental relation \( G^2 = A \cdot H \) for two positive numbers. This formula is a quick way to find any of the three means if the other two are known.

Question 10. If \( S_n \) denote the sum of \( n \) terms of an A. P whose common difference is \( d \), the value of \( S_n – 2S_{n-1} + S_{n-2} \) is
(a) \( d \)
(b) \( d^2 \)
Answer: (a) \( d \)
In simple words: We can break down the given expression by understanding that the difference between consecutive sums in an Arithmetic Progression gives the term itself. The term \( T_n = S_n - S_{n-1} \). Using this, the expression simplifies to the difference between two consecutive terms, which is the common difference, \( d \). This shows how the sum formula directly relates to the terms and common difference.
Let \( T_k \) be the \( k \)-th term of the A.P. We know that \( T_k = S_k - S_{k-1} \).
The given expression is \( S_n - 2S_{n-1} + S_{n-2} \).
We can rewrite this as:
\( (S_n - S_{n-1}) - (S_{n-1} - S_{n-2}) \)
Using the definition \( T_k = S_k - S_{k-1} \):
\( = T_n - T_{n-1} \)
For an Arithmetic Progression, the difference between consecutive terms is the common difference \( d \).
\( T_n = a + (n-1)d \)
\( T_{n-1} = a + (n-1-1)d = a + (n-2)d \)
So, \( T_n - T_{n-1} = (a + (n-1)d) - (a + (n-2)d) \)
\( = a + nd - d - a - nd + 2d \)
\( = d \)

🎯 Exam Tip: Remember the relation \( T_n = S_n - S_{n-1} \). This identity is very useful for solving problems involving sums and terms of sequences. Also, \( T_n - T_{n-1} = d \) for an A.P.

Question 11. The remainder when \( 38^{15} \) is divided by \( 13 \) is
(a) \( 12 \)
(b) (Missing option text)
(c) \( 11 \)
(d) \( 5 \)
Answer: (a) \( 12 \)
In simple words: To find the remainder of \( 38^{15} \) divided by \( 13 \), we can use the binomial theorem. We write \( 38 \) as \( (39-1) \), and since \( 39 \) is a multiple of \( 13 \), most terms in the expansion will be divisible by \( 13 \). The remainder will come from the last term. This method simplifies the calculation significantly.
We need to find the remainder of \( 38^{15} \div 13 \).
We can write \( 38 \) as \( 39 - 1 \). Since \( 39 \) is a multiple of \( 13 \) (i.e., \( 39 = 3 \times 13 \)), we have:
\( 38^{15} = (39 - 1)^{15} \)
Using the binomial expansion of \( (a-b)^n \):
\( (39 - 1)^{15} = 39^{15} - 15C_1 39^{14}(1) + 15C_2 39^{13}(1)^2 - \dots + 15C_{14} 39^1(1)^{14} - 15C_{15}(1)^{15} \)
All terms containing \( 39 \) as a factor are divisible by \( 13 \).
So, \( 39^{15} \equiv 0 \pmod{13} \)
\( 15C_1 39^{14}(1) \equiv 0 \pmod{13} \)
...and so on for all terms except the last one, \( -15C_{15}(1)^{15} \).
The last term is \( -15C_{15}(1)^{15} = -1 \cdot 1 = -1 \).
So, \( 38^{15} \equiv -1 \pmod{13} \).
A remainder cannot be negative. To find the positive remainder, add the divisor:
Remainder \( = -1 + 13 = 12 \).

🎯 Exam Tip: For remainder problems, try to express the base as a multiple of the divisor plus or minus a small number (e.g., \( 38 = 3 \times 13 - 1 \) or \( 38 = 2 \times 13 + 12 \)). The binomial theorem helps expand this for powers.

Question 12. The nth term of the sequence \( 1, 2, 4, 7, 11, \dots \) is
(a) \( n^2 + 3n^2 + 2n \)
(b) \( n^3 – 3n^2 + 3n \)
(c) \( \frac{n(n+1)(n+2)}{3} \)
(d) \( \frac{n^{2}-n+2}{2} \)
Answer: (d) \( \frac{n^{2}-n+2}{2} \)
In simple words: This is a sequence where the differences between consecutive terms form another sequence. We check if the given formula for the \( n \)-th term correctly generates the terms of the sequence. By plugging in \( n=1, 2, 3, 4 \) into the formula, we can verify that it produces the terms \( 1, 2, 4, 7 \) respectively.
Let's test the given formula \( a_n = \frac{n^2 - n + 2}{2} \):
For \( n=1 \): \( a_1 = \frac{1^2 - 1 + 2}{2} = \frac{1 - 1 + 2}{2} = \frac{2}{2} = 1 \)
For \( n=2 \): \( a_2 = \frac{2^2 - 2 + 2}{2} = \frac{4 - 2 + 2}{2} = \frac{4}{2} = 2 \)
For \( n=3 \): \( a_3 = \frac{3^2 - 3 + 2}{2} = \frac{9 - 3 + 2}{2} = \frac{8}{2} = 4 \)
For \( n=4 \): \( a_4 = \frac{4^2 - 4 + 2}{2} = \frac{16 - 4 + 2}{2} = \frac{14}{2} = 7 \)
The terms match the given sequence \( 1, 2, 4, 7, \dots \).
Therefore, the \( n \)-th term is \( \frac{n^{2}-n+2}{2} \).

🎯 Exam Tip: For sequences where the \( n \)-th term formula is given in options, the easiest way to verify is to substitute \( n=1, 2, 3 \) and check if the terms match the sequence.

Question 13. The sum up to \( n \) terms of the series \( \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots \) is
(a) \( \sqrt{2 n+1} \)
(b) \( \frac{\sqrt{2 n+1}}{2} \)
(c) \( \sqrt{2 n+1}-1 \)
(d) \( \frac{\sqrt{2 n+1}-1}{2} \)
Answer: (d) \( \frac{\sqrt{2 n+1}-1}{2} \)
In simple words: This series involves square roots in the denominator. To simplify each term, we rationalize the denominator by multiplying the numerator and denominator by the conjugate. This process leads to a telescoping series, where most of the terms cancel out, leaving only the first and last parts.
The \( n \)-th term of the series is \( T_n = \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} \).
Rationalize the denominator of \( T_n \):
\( T_n = \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} \times \frac{\sqrt{2n-1} - \sqrt{2n+1}}{\sqrt{2n-1} - \sqrt{2n+1}} \)
\( T_n = \frac{\sqrt{2n-1} - \sqrt{2n+1}}{(2n-1) - (2n+1)} \)
\( T_n = \frac{\sqrt{2n-1} - \sqrt{2n+1}}{-2} \)
\( T_n = \frac{\sqrt{2n+1} - \sqrt{2n-1}}{2} \)
Now, let's write out the first few terms and the \( n \)-th term for the sum \( S_n \):
\( T_1 = \frac{\sqrt{3} - \sqrt{1}}{2} \)
\( T_2 = \frac{\sqrt{5} - \sqrt{3}}{2} \)
\( T_3 = \frac{\sqrt{7} - \sqrt{5}}{2} \)
...
\( T_n = \frac{\sqrt{2n+1} - \sqrt{2n-1}}{2} \)
Summing these terms \( S_n = T_1 + T_2 + \dots + T_n \):
\( S_n = \frac{1}{2} [(\sqrt{3} - \sqrt{1}) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \dots + (\sqrt{2n+1} - \sqrt{2n-1})] \)
All intermediate terms cancel out (telescoping sum).
\( S_n = \frac{1}{2} [\sqrt{2n+1} - \sqrt{1}] \)
\( S_n = \frac{\sqrt{2n+1} - 1}{2} \)

🎯 Exam Tip: For series involving square roots in the denominator, always try to rationalize each term first. This often leads to a telescoping sum where most terms cancel out, making the sum easy to calculate.

Question 14. The nth term of the sequence \( \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \dots \) is
(a) \( 2^n - n - 1 \)
(b) \( 1 – 2^{-n} \)
(c) \( 2^{-n} + n – 1 \)
(d) \( 2^{n-1} \)
Answer: (b) \( 1 – 2^{-n} \)
In simple words: Observe the pattern of each term. The denominators are powers of 2. The numerators are one less than the denominators. For example, the first term is \( \frac{2^1-1}{2^1} \), the second is \( \frac{2^2-1}{2^2} \), and so on. This pattern can be written as \( \frac{2^n-1}{2^n} \), which simplifies to \( 1 - \frac{1}{2^n} \) or \( 1 - 2^{-n} \).
Let's analyze the terms of the sequence:
\( T_1 = \frac{1}{2} = \frac{2^1-1}{2^1} \)
\( T_2 = \frac{3}{4} = \frac{2^2-1}{2^2} \)
\( T_3 = \frac{7}{8} = \frac{2^3-1}{2^3} \)
\( T_4 = \frac{15}{16} = \frac{2^4-1}{2^4} \)
Following this pattern, the \( n \)-th term \( T_n \) is:
\( T_n = \frac{2^n-1}{2^n} \)
This can be rewritten as:
\( T_n = \frac{2^n}{2^n} - \frac{1}{2^n} \)
\( T_n = 1 - \frac{1}{2^n} \)
\( T_n = 1 - 2^{-n} \)

🎯 Exam Tip: For sequences involving fractions, check if the numerators and denominators follow separate patterns, especially powers or arithmetic progressions. Look for a relation between the numerator and denominator for each term.

Question 15. The sum up to \( n \) terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots \) is
(a) \( \frac{n(n+1)}{2} \)
(b) \( 2n (n + 1) \)
(c) \( \frac{n(n+1)}{\sqrt{2}} \)
(d) \( 1 \)
Answer: (c) \( \frac{n(n+1)}{\sqrt{2}} \)
In simple words: First, simplify each term in the series by taking out perfect square factors from inside the square roots. You'll notice that each term becomes a multiple of \( \sqrt{2} \). Then, factor out \( \sqrt{2} \) from the entire series, which will leave a simple arithmetic sum of natural numbers. Finally, use the formula for the sum of natural numbers to find the total sum.
The given series is \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots \)
Simplify each term:
\( \sqrt{2} = 1\sqrt{2} \)
\( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
\( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \)
\( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)
So the series becomes: \( 1\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} + \dots \)
Factor out \( \sqrt{2} \):
\( S_n = \sqrt{2} (1 + 2 + 3 + 4 + \dots + n) \)
The sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \).
So, \( S_n = \sqrt{2} \left( \frac{n(n+1)}{2} \right) \)
To match the option, we can write \( \sqrt{2} \) as \( \frac{2}{\sqrt{2}} \):
\( S_n = \frac{2}{\sqrt{2}} \left( \frac{n(n+1)}{2} \right) \)
\( S_n = \frac{n(n+1)}{\sqrt{2}} \)

🎯 Exam Tip: Always simplify square root terms first. Recognizing common factors or patterns will often transform complex series into simpler, known sums (like arithmetic or geometric series).

Question 16. The value of the series \( \frac{1}{2} + \frac{7}{4} + \frac{13}{8} + \frac{19}{16} + \dots \) is
(a) \( 14 \)
(b) \( 7 \)
(c) \( 4 \)
(d) \( 6 \)
Answer: (a) \( 14 \)
In simple words: This is an Arithmetico-Geometric Progression (AGP) which is a combination of an A.P. and a G.P. The numerators \( 1, 7, 13, 19, \dots \) form an A.P. with \( a=1, d=6 \). The denominators \( 2, 4, 8, 16, \dots \) show a G.P. common ratio \( r=1/2 \) in the fraction. The sum of an infinite AGP can be found using a specific formula.
The given series is \( S_\infty = \frac{1}{2} + \frac{7}{4} + \frac{13}{8} + \frac{19}{16} + \dots \)
This is an Arithmetico-Geometric Progression.
The A.P. part has first term \( a=1 \) and common difference \( d=6 \) (from numerators: \( 1, 7, 13, 19, \dots \)).
The G.P. part has first term \( x=1/2 \) and common ratio \( r=1/2 \) (from \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots \)).
The sum of an infinite AGP is given by the formula:
\( S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} \)
Substitute the values: \( a=1 \), \( d=6 \), \( r=1/2 \).
\( S_\infty = \frac{1}{1 - \frac{1}{2}} + \frac{6 \cdot \frac{1}{2}}{(1 - \frac{1}{2})^2} \)
\( S_\infty = \frac{1}{\frac{1}{2}} + \frac{3}{(\frac{1}{2})^2} \)
\( S_\infty = 2 + \frac{3}{\frac{1}{4}} \)
\( S_\infty = 2 + 3 \times 4 \)
\( S_\infty = 2 + 12 \)
\( S_\infty = 14 \)

🎯 Exam Tip: Identify if a series is an AGP. If it is, know the formula for its sum to infinity. Remember that the formula is valid only when \( |r| < 1 \).

Question 17. The sum of an infinite G.P is 18. If the first term is 6 the common ratio is
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{3}{4} \)
Answer: (b) \( \frac{2}{3} \)
In simple words: The sum of an infinite Geometric Progression (G.P.) has a simple formula if its common ratio is between -1 and 1. We are given the sum and the first term, so we can use this formula to calculate the common ratio.
Given: Sum of infinite G.P. \( S_\infty = 18 \)
First term \( a = 6 \)
The formula for the sum of an infinite G.P. is \( S_\infty = \frac{a}{1-r} \), where \( r \) is the common ratio.
Substitute the given values into the formula:
\( 18 = \frac{6}{1-r} \)
Multiply both sides by \( (1-r) \):
\( 18(1-r) = 6 \)
Divide both sides by \( 18 \):
\( 1-r = \frac{6}{18} \)
\( 1-r = \frac{1}{3} \)
Now, solve for \( r \):
\( r = 1 - \frac{1}{3} \)
\( r = \frac{3-1}{3} \)
\( r = \frac{2}{3} \)

🎯 Exam Tip: Make sure to correctly rearrange the formula \( S_\infty = \frac{a}{1-r} \) to solve for \( r \). Always check that the calculated common ratio \( r \) satisfies \( |r| < 1 \) for the sum to infinity to exist.

Question 18. The coefficient of \( x^5 \) in the series \( e^{-2x} \) is
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{2} \)
(c) \( -\frac{4}{15} \)
(d) \( \frac{4}{15} \)
Answer: (c) \( -\frac{4}{15} \)
In simple words: The exponential function \( e^u \) has a known series expansion. We substitute \( u = -2x \) into this general expansion. Then, we look for the term that contains \( x^5 \) and identify its coefficient.
The Maclaurin series expansion for \( e^u \) is given by:
\( e^u = 1 + \frac{u}{1!} + \frac{u^2}{2!} + \frac{u^3}{3!} + \frac{u^4}{4!} + \frac{u^5}{5!} + \dots \)
In this problem, \( u = -2x \). Substitute this into the expansion:
\( e^{-2x} = 1 + \frac{(-2x)}{1!} + \frac{(-2x)^2}{2!} + \frac{(-2x)^3}{3!} + \frac{(-2x)^4}{4!} + \frac{(-2x)^5}{5!} + \dots \)
We are interested in the coefficient of \( x^5 \), which comes from the term \( \frac{(-2x)^5}{5!} \).
\( \frac{(-2x)^5}{5!} = \frac{(-2)^5 x^5}{5!} = \frac{-32 x^5}{120} \)
So the coefficient of \( x^5 \) is \( \frac{-32}{120} \).
Simplify the fraction:
\( \frac{-32}{120} = \frac{-4 \times 8}{15 \times 8} = -\frac{4}{15} \)

🎯 Exam Tip: Remember the standard Maclaurin series for \( e^x \). Be careful with negative signs and factorials when calculating coefficients for specific powers.

Question 19. The value of \( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \) is
(a) \( \frac{e^{2}+1}{2 e} \)
(b) \( \frac{(e+1)^{2}}{2 e} \)
(c) \( \frac{(e-1)^{2}}{2 e} \)
Answer: (c) \( \frac{(e-1)^{2}}{2 e} \)
In simple words: This series involves only even factorials. We can derive its sum by using the known Maclaurin series expansions for \( e^x \) and \( e^{-x} \). By adding these two series and rearranging terms, we can isolate the sum of terms with even factorials.
The Maclaurin series for \( e^x \) is:
\( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \)
The Maclaurin series for \( e^{-x} \) is:
\( e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots \)
Add the two series:
\( e^x + e^{-x} = 2 + 2\frac{x^2}{2!} + 2\frac{x^4}{4!} + 2\frac{x^6}{6!} + \dots \)
Divide by 2:
\( \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots \)
Now, substitute \( x=1 \) into this equation:
\( \frac{e^1 + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \)
The series we want to sum is \( S = \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \)
From the above, \( S = \frac{e + e^{-1}}{2} - 1 \)
Now, simplify this expression:
\( S = \frac{e + \frac{1}{e}}{2} - 1 \)
\( S = \frac{\frac{e^2 + 1}{e}}{2} - 1 \)
\( S = \frac{e^2 + 1}{2e} - 1 \)
\( S = \frac{e^2 + 1 - 2e}{2e} \)
\( S = \frac{(e-1)^2}{2e} \)

🎯 Exam Tip: To find sums of series involving only even or odd factorials, use the series expansions of \( e^x \) and \( e^{-x} \). Adding them isolates even powers, and subtracting them isolates odd powers.

Question 20. The value of \( 1 - \frac{1}{2}(\frac{2}{3}) + \frac{1}{3}(\frac{2}{3})^2 - \frac{1}{4}(\frac{2}{3})^3 + \dots \) is
(a) \( \log \left(\frac{5}{3}\right) \)
(b) \( \frac{3}{2} \log \left(\frac{5}{3}\right) \)
(c) \( \frac{5}{3} \log \left(\frac{5}{3}\right) \)
(d) \( \frac{2}{3} \log \left(\frac{2}{3}\right) \)
Answer: (b) \( \frac{3}{2} \log \left(\frac{5}{3}\right) \)
In simple words: The series given is related to the logarithmic expansion \( \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \). We need to manipulate the given series to match this standard form. By multiplying the entire series by a specific factor, we can transform it into the expansion of \( \log(1+x) \) for a particular \( x \).
Let the given series be \( S \):
\( S = 1 - \frac{1}{2}\left(\frac{2}{3}\right) + \frac{1}{3}\left(\frac{2}{3}\right)^2 - \frac{1}{4}\left(\frac{2}{3}\right)^3 + \dots \)
The standard Maclaurin series for \( \log(1+y) \) is:
\( \log(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \frac{y^4}{4} + \dots \)
Let's multiply the series \( S \) by \( \frac{2}{3} \):
\( \frac{2}{3} S = \frac{2}{3} \cdot 1 - \frac{2}{3} \cdot \frac{1}{2}\left(\frac{2}{3}\right) + \frac{2}{3} \cdot \frac{1}{3}\left(\frac{2}{3}\right)^2 - \frac{2}{3} \cdot \frac{1}{4}\left(\frac{2}{3}\right)^3 + \dots \)
\( \frac{2}{3} S = \frac{2}{3} - \frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{3}\left(\frac{2}{3}\right)^3 - \frac{1}{4}\left(\frac{2}{3}\right)^4 + \dots \)
This new series is exactly the expansion for \( \log(1+y) \) with \( y = \frac{2}{3} \).
So, \( \frac{2}{3} S = \log\left(1 + \frac{2}{3}\right) \)
\( \frac{2}{3} S = \log\left(\frac{3+2}{3}\right) \)
\( \frac{2}{3} S = \log\left(\frac{5}{3}\right) \)
Now, solve for \( S \):
\( S = \frac{3}{2} \log\left(\frac{5}{3}\right) \)

🎯 Exam Tip: When given a non-standard series, compare it to known series expansions like those for \( \log(1+x) \), \( e^x \), \( \sin x \), etc. Sometimes a simple multiplication or differentiation/integration can transform it into a recognizable form.

Question 20. The value of \( 1-\frac{1}{2}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^{2}-\frac{1}{4}\left(\frac{2}{3}\right)^{3}+\ldots \ldots \ldots \) is
(1) \( \log \left(\frac{5}{3}\right) \)
(2) \( \frac{3}{2} \log \left(\frac{5}{3}\right) \)
(3) \( \frac{5}{3} \log \left(\frac{5}{3}\right) \)
(4) \( \frac{2}{3} \log \left(\frac{2}{3}\right) \)
Answer: (2) \( \frac{3}{2} \log \left(\frac{5}{3}\right) \)
In simple words: This question asks for the sum of a special type of series. We can compare it to the known series expansion of the logarithm function to find its value.

🎯 Exam Tip: Recognizing standard series expansions like those for \( \log(1+x) \) or \( e^x \) is crucial for quickly solving such problems. Practice helps in identifying the correct form and substituting the values accurately.

Explaination:
The given series is \( S = 1-\frac{1}{2}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^{2}-\frac{1}{4}\left(\frac{2}{3}\right)^{3}+\ldots \ldots \ldots \)
We know the expansion for \( \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \)
To match the given series with the logarithmic expansion, we can multiply the given series S by \( \frac{2}{3} \).
So, let \( S = 1-\frac{1}{2}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^{2}-\frac{1}{4}\left(\frac{2}{3}\right)^{3}+\ldots \)
Multiplying both sides by \( \frac{2}{3} \):
\( \frac{2}{3} S = \frac{2}{3} - \frac{1}{2}\left(\frac{2}{3}\right)^{2}+\frac{1}{3}\left(\frac{2}{3}\right)^{3}-\frac{1}{4}\left(\frac{2}{3}\right)^{4}+\ldots \)
This new series is exactly the expansion of \( \log(1+x) \) where \( x = \frac{2}{3} \). This is a standard result in calculus.
Therefore,
\( \frac{2}{3} S = \log\left(1 + \frac{2}{3}\right) \)

\( \implies \frac{2}{3} S = \log\left(\frac{3+2}{3}\right) \)

\( \implies \frac{2}{3} S = \log\left(\frac{5}{3}\right) \)
Now, we solve for S:
\( S = \frac{3}{2} \log\left(\frac{5}{3}\right) \)
In simple words: We changed the given series by multiplying it by a number so it looks like a known logarithm series. After that, we found the sum using the logarithm formula and then worked backward to find the original series's value.

🎯 Exam Tip: Always remember that the general term of a series can help you identify its type (arithmetic, geometric, or a known function expansion like Taylor series). Sometimes, a small algebraic manipulation is needed to transform the given series into a recognizable form.