RBSE Solutions Class 11 Maths Chapter 8 Sequence, Progression, and Series Exercise 8.6

Question 1. Find the sum of n term of that series whose nth term is:
(i) \( 3n^2 + 2n + 5 \)
(ii) \( 4n^3 + 7n + 1 \)
(iii) \( n(n + 1)(n + 2) \)
Answer:
(i) Given, the nth term \( T_n = 3n^2 + 2n + 5 \).
To find the sum of n terms, \( S_n = \Sigma T_n \). We use the standard summation formulas:
\( \Sigma n = \frac{n(n+1)}{2} \)
\( \Sigma n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \Sigma n^3 = \left[\frac{n(n+1)}{2}\right]^2 \)
Now, we apply these formulas to find \( S_n \):
\( S_n = \Sigma (3n^2 + 2n + 5) \)
\( S_n = 3\Sigma n^2 + 2\Sigma n + 5\Sigma 1 \)
\( S_n = 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} + 5n \)
\( S_n = \frac{n(n+1)(2n+1)}{2} + n(n+1) + 5n \)
To simplify the expression, we can take common factors:
\( S_n = n(n+1) \left[ \frac{2n+1}{2} + 1 \right] + 5n \)
\( S_n = n(n+1) \left[ \frac{2n+1+2}{2} \right] + 5n \)
\( S_n = \frac{n(n+1)(2n+3)}{2} + 5n \)
This is the sum of n terms for the first series. Understanding these basic summation formulas helps solve complex series problems efficiently.
(ii) Given, the nth term \( T_n = 4n^3 + 7n + 1 \).
The sum of n terms is \( S_n = \Sigma T_n \). We apply the summation formulas:
\( S_n = \Sigma (4n^3 + 7n + 1) \)
\( S_n = 4\Sigma n^3 + 7\Sigma n + \Sigma 1 \)
\( S_n = 4 \left[ \frac{n(n+1)}{2} \right]^2 + 7 \cdot \frac{n(n+1)}{2} + n \)
\( S_n = 4 \cdot \frac{n^2(n+1)^2}{4} + 7 \cdot \frac{n(n+1)}{2} + n \)
\( S_n = n^2(n+1)^2 + 7 \cdot \frac{n(n+1)}{2} + n \)
Now, we take common factors to simplify:
\( S_n = n(n+1) \left[ n(n+1) + \frac{7}{2} \right] + n \)
\( S_n = n(n+1) \left[ \frac{2n(n+1) + 7}{2} \right] + n \)
\( S_n = n(n+1) \left[ \frac{2n^2 + 2n + 7}{2} \right] + n \)
\( S_n = \frac{n(n+1)(2n^2 + 2n + 7)}{2} + n \)
This provides the sum of n terms for the second series. Series summation is a fundamental concept in mathematics that helps find the total value of many numbers quickly.
(iii) Given, the nth term \( T_n = n(n + 1)(n + 2) \).
First, we expand the expression for \( T_n \):
\( T_n = n(n^2 + 3n + 2) \)
\( T_n = n^3 + 3n^2 + 2n \)
Now, we find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (n^3 + 3n^2 + 2n) \)
\( S_n = \Sigma n^3 + 3\Sigma n^2 + 2\Sigma n \)
Apply the summation formulas:
\( S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} \)
\( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) \)
To simplify, take \( \frac{n(n+1)}{4} \) as a common factor:
\( S_n = \frac{n(n+1)}{4} \left[ n(n+1) + 2(2n+1) + 4 \right] \)
\( S_n = \frac{n(n+1)}{4} [n^2 + n + 4n + 2 + 4] \)
\( S_n = \frac{n(n+1)}{4} [n^2 + 5n + 6] \)
Factor the quadratic expression \( n^2 + 5n + 6 \). It factors into \( (n+2)(n+3) \):
\( S_n = \frac{n(n+1)(n+2)(n+3)}{4} \)
This gives the sum of n terms for the third series. Recognizing patterns and simplifying algebraic expressions is key in solving such problems.
In simple words: To find the sum of a series, we use special formulas for summing up numbers, squares, and cubes up to 'n' terms. We substitute the nth term into the sum formula and then simplify the expression by combining like terms and factoring.

🎯 Exam Tip: Remember the basic summation formulas for \( \Sigma n \), \( \Sigma n^2 \), and \( \Sigma n^3 \) as they are essential building blocks for solving these types of problems quickly and accurately.

Question 2. Find the sum of n terms of following series:
(i) \( 3^2 + 7^2 + 11^2 + 15^2 + ... \)
(ii) \( 2^3 + 5^3 + 8^3 + 11^3 + ... \)
(iii) \( 1.2^2 + 2.3^2 + 3.4^2 + ... \)
Answer:
(i) Given series: \( 3^2 + 7^2 + 11^2 + 15^2 + ... \)
First, find the nth term \( T_n \). The bases (3, 7, 11, 15, ...) form an Arithmetic Progression (AP) with first term \( a = 3 \) and common difference \( d = 4 \).
The nth term of this AP is \( a_n = a + (n-1)d = 3 + (n-1)4 = 3 + 4n - 4 = 4n - 1 \).
So, the nth term of the given series is \( T_n = (4n - 1)^2 \).
Expand \( T_n \): \( T_n = (4n)^2 - 2(4n)(1) + 1^2 = 16n^2 - 8n + 1 \).
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (16n^2 - 8n + 1) \)
\( S_n = 16\Sigma n^2 - 8\Sigma n + \Sigma 1 \)
Apply the summation formulas:
\( S_n = 16 \cdot \frac{n(n+1)(2n+1)}{6} - 8 \cdot \frac{n(n+1)}{2} + n \)
\( S_n = \frac{8n(n+1)(2n+1)}{3} - 4n(n+1) + n \)
Take \( n \) as a common factor:
\( S_n = n \left[ \frac{8(n+1)(2n+1)}{3} - 4(n+1) + 1 \right] \)
\( S_n = \frac{n}{3} \left[ 8(n+1)(2n+1) - 12(n+1) + 3 \right] \)
\( S_n = \frac{n}{3} \left[ 8(2n^2 + 3n + 1) - 12n - 12 + 3 \right] \)
\( S_n = \frac{n}{3} [16n^2 + 24n + 8 - 12n - 9] \)
\( S_n = \frac{n}{3} [16n^2 + 12n - 1] \)
This is the sum of n terms for the first series. Finding the nth term correctly is the first and most crucial step in solving these series problems.
(iii) Given series: \( 1.2^2 + 2.3^2 + 3.4^2 + ... \)
First, find the nth term \( T_n \).
The first part of each term (1, 2, 3, ...) is simply \( n \).
The second part of each term (\( 2^2, 3^2, 4^2, ... \)) is \( (n+1)^2 \).
So, the nth term of the given series is \( T_n = n(n+1)^2 \).
Expand \( T_n \): \( T_n = n(n^2 + 2n + 1) = n^3 + 2n^2 + n \).
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (n^3 + 2n^2 + n) \)
\( S_n = \Sigma n^3 + 2\Sigma n^2 + \Sigma n \)
Apply the summation formulas:
\( S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \)
\( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \)
To simplify, take \( \frac{n(n+1)}{12} \) as a common factor (LCM of 4, 3, 2 is 12):
\( S_n = \frac{n(n+1)}{12} \left[ 3n(n+1) + 4(2n+1) + 6 \right] \)
\( S_n = \frac{n(n+1)}{12} [3n^2 + 3n + 8n + 4 + 6] \)
\( S_n = \frac{n(n+1)}{12} [3n^2 + 11n + 10] \)
Factor the quadratic expression \( 3n^2 + 11n + 10 \). It factors into \( (n+2)(3n+5) \):
\( S_n = \frac{n(n+1)(n+2)(3n+5)}{12} \)
This is the sum of n terms for the third series. Breaking down complex terms into simpler patterns makes it easier to apply known formulas.
In simple words: For series where terms are squares or products, first find a pattern to write the 'nth' term. Then, use the sum formulas for n, n-squared, and n-cubed to add them all up.

🎯 Exam Tip: Always double-check your nth term derivation. A small error in the nth term will lead to a completely incorrect sum of the series.

Question 3. Find the nth term and sum of n terms of following series:
(i) \( 1.3+3.5+ 5.7 + ... \)
(ii) \( 1.2.4 + 2.3.7 + 3.4.10 + ... \)
Answer:
(i) Given series: \( 1.3 + 3.5 + 5.7 + ... \)
First, find the nth term \( T_n \). Each term is a product of two numbers.
The first parts of the products (1, 3, 5, ...) form an Arithmetic Progression (AP) with first term \( a = 1 \) and common difference \( d = 2 \). Its nth term is \( a_n = 1 + (n-1)2 = 1 + 2n - 2 = 2n - 1 \).
The second parts of the products (3, 5, 7, ...) form an AP with first term \( a = 3 \) and common difference \( d = 2 \). Its nth term is \( a_n = 3 + (n-1)2 = 3 + 2n - 2 = 2n + 1 \).
So, the nth term of the given series is \( T_n = (2n - 1)(2n + 1) \).
Expand \( T_n \): \( T_n = (2n)^2 - 1^2 = 4n^2 - 1 \).
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (4n^2 - 1) \)
\( S_n = 4\Sigma n^2 - \Sigma 1 \)
Apply the summation formulas:
\( S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} - n \)
\( S_n = \frac{2n(n+1)(2n+1)}{3} - n \)
Take \( n \) as a common factor:
\( S_n = n \left[ \frac{2(n+1)(2n+1)}{3} - 1 \right] \)
\( S_n = \frac{n}{3} [2(n+1)(2n+1) - 3] \)
\( S_n = \frac{n}{3} [2(2n^2 + 3n + 1) - 3] \)
\( S_n = \frac{n}{3} [4n^2 + 6n + 2 - 3] \)
\( S_n = \frac{n}{3} [4n^2 + 6n - 1] \)
This gives the nth term and the sum of n terms for the first series. Breaking down the product into individual APs is a clever way to find the general term.
(ii) Given series: \( 1.2.4 + 2.3.7 + 3.4.10 + ... \)
First, find the nth term \( T_n \). Each term is a product of three numbers.
The first parts (1, 2, 3, ...) form an AP with nth term \( n \).
The second parts (2, 3, 4, ...) form an AP with nth term \( n+1 \).
The third parts (4, 7, 10, ...) form an AP with first term \( a = 4 \) and common difference \( d = 3 \). Its nth term is \( a_n = 4 + (n-1)3 = 4 + 3n - 3 = 3n + 1 \).
So, the nth term of the given series is \( T_n = n(n+1)(3n+1) \).
Expand \( T_n \):
\( T_n = (n^2 + n)(3n + 1) \)
\( T_n = 3n^3 + n^2 + 3n^2 + n \)
\( T_n = 3n^3 + 4n^2 + n \)
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (3n^3 + 4n^2 + n) \)
\( S_n = 3\Sigma n^3 + 4\Sigma n^2 + \Sigma n \)
Apply the summation formulas:
\( S_n = 3 \left[ \frac{n(n+1)}{2} \right]^2 + 4 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \)
\( S_n = \frac{3n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \)
To simplify, take \( \frac{n(n+1)}{12} \) as a common factor (LCM of 4, 3, 2 is 12):
\( S_n = \frac{n(n+1)}{12} \left[ 9n(n+1) + 8(2n+1) + 6 \right] \)
\( S_n = \frac{n(n+1)}{12} [9n^2 + 9n + 16n + 8 + 6] \)
\( S_n = \frac{n(n+1)}{12} [9n^2 + 25n + 14] \)
Factor the quadratic expression \( 9n^2 + 25n + 14 \). It factors into \( (n+2)(9n+7) \):
\( S_n = \frac{n(n+1)(n+2)(9n+7)}{12} \)
This gives the nth term and the sum of n terms for the second series. Carefully breaking down each factor in the term into an AP simplifies the process significantly.
In simple words: For series with terms that are products, figure out the pattern for each factor to find the 'nth' term. Then, multiply them out and use the sum formulas for n, n-squared, and n-cubed to get the total sum.

🎯 Exam Tip: When dealing with product series, focus on finding the general term for each factor separately before multiplying them to get the complete nth term.

Question 4. Find the nth term and sum of n terms of following series:
(i) \( 3+8+15+ 24 + ... \)
(ii) \( 1+6+13+22 + ... \)
Answer:
(i) Given series: \( 3+8+15+ 24 + ... \)
Let the sum of n terms be \( S_n = 3+8+15+24+...+T_n \).
Write \( S_n \) again, shifting terms by one position:
\( S_n = 0 + 3+8+15+...+T_{n-1}+T_n \)
Subtracting the second equation from the first:
\( 0 = 3 + (8-3) + (15-8) + (24-15) + ... + (T_n - T_{n-1}) - T_n \)
\( 0 = 3 + 5 + 7 + 9 + ... \text{ (n terms)} - T_n \)
This means \( T_n = 3 + 5 + 7 + 9 + ... \text{ (n terms)} \).
The sequence \( 3, 5, 7, 9, ... \) is an AP with first term \( a = 3 \) and common difference \( d = 2 \).
The sum of n terms of an AP is given by \( \frac{n}{2}[2a + (n-1)d] \).
So, \( T_n = \frac{n}{2}[2(3) + (n-1)2] \)
\( T_n = \frac{n}{2}[6 + 2n - 2] \)
\( T_n = \frac{n}{2}[2n + 4] \)
\( T_n = n(n+2) \)
\( T_n = n^2 + 2n \)
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (n^2 + 2n) \)
\( S_n = \Sigma n^2 + 2\Sigma n \)
Apply the summation formulas:
\( S_n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} \)
\( S_n = \frac{n(n+1)(2n+1)}{6} + n(n+1) \)
Take \( \frac{n(n+1)}{6} \) as a common factor:
\( S_n = \frac{n(n+1)}{6} [(2n+1) + 6] \)
\( S_n = \frac{n(n+1)(2n+7)}{6} \)
This method is called the method of differences, useful when the differences between consecutive terms form an AP.
(ii) Given series: \( 1+6+13+22 + ... \)
Let the terms be \( T_1, T_2, T_3, ... \).
\( T_1 = 1 \)
\( T_2 = 6 \)
\( T_3 = 13 \)
\( T_4 = 22 \)
Let's find the differences between consecutive terms:
First differences: \( 6-1=5, 13-6=7, 22-13=9, ... \)
The sequence of first differences \( (5, 7, 9, ...) \) is an AP with first term \( a'=5 \) and common difference \( d'=2 \).
Since the first differences are in AP, the nth term \( T_n \) of the original series will be a quadratic expression in 'n' of the form \( An^2 + Bn + C \).
For \( n=1 \): \( A(1)^2 + B(1) + C = 1 \implies A+B+C = 1 \)
For \( n=2 \): \( A(2)^2 + B(2) + C = 6 \implies 4A+2B+C = 6 \)
For \( n=3 \): \( A(3)^2 + B(3) + C = 13 \implies 9A+3B+C = 13 \)
Subtracting the first equation from the second, we get: \( (4A+2B+C) - (A+B+C) = 6 - 1 \implies 3A+B = 5 \).
Subtracting the second equation from the third, we get: \( (9A+3B+C) - (4A+2B+C) = 13 - 6 \implies 5A+B = 7 \).
Now, subtract \( (3A+B=5) \) from \( (5A+B=7) \): \( (5A+B) - (3A+B) = 7 - 5 \implies 2A = 2 \implies A = 1 \).
Substitute \( A=1 \) into \( 3A+B=5 \): \( 3(1)+B=5 \implies B = 2 \).
Substitute \( A=1 \) and \( B=2 \) into \( A+B+C=1 \): \( 1+2+C=1 \implies 3+C=1 \implies C = -2 \).
So, the nth term is \( T_n = n^2 + 2n - 2 \).
Now, find the sum of n terms, \( S_n = \Sigma T_n \):
\( S_n = \Sigma (n^2 + 2n - 2) \)
\( S_n = \Sigma n^2 + 2\Sigma n - 2\Sigma 1 \)
Apply the summation formulas:
\( S_n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} - 2n \)
\( S_n = \frac{n(n+1)(2n+1)}{6} + n(n+1) - 2n \)
Take \( \frac{n}{6} \) as a common factor:
\( S_n = \frac{n}{6} [(n+1)(2n+1) + 6(n+1) - 12] \)
\( S_n = \frac{n}{6} [2n^2 + 3n + 1 + 6n + 6 - 12] \)
\( S_n = \frac{n}{6} [2n^2 + 9n - 5] \)
Factor the quadratic expression \( 2n^2 + 9n - 5 \). It factors into \( (2n-1)(n+5) \):
\( S_n = \frac{n(2n-1)(n+5)}{6} \)
This method of finding the nth term from differences is robust for polynomial sequences.
In simple words: When terms don't follow a direct formula, we look at the differences between terms. If these differences form a pattern (like an AP), we can find the general nth term. Once we have the nth term, we sum it up using standard summation formulas.

🎯 Exam Tip: For sequences where the first differences are in an AP, assume the general term is a quadratic in 'n' ( \( An^2 + Bn + C \) ) and solve for A, B, C using initial terms. For sequences where the second differences are in AP, assume a cubic general term, and so on.

Question 5. Find the nth term and sum of n terms of the following series:
(i) \( 1 + (1 + 2) + (1 + 2 + 3) + ... \)
(ii) \( 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... \)
Answer:
(i) Given series: \( 1 + (1 + 2) + (1 + 2 + 3) + ... \)
The nth term \( T_n \) of this series is the sum of the first n natural numbers:
\( T_n = 1 + 2 + 3 + ... + n \)
Using the formula for the sum of the first n natural numbers:
\( T_n = \frac{n(n+1)}{2} \)
Now, find the sum of n terms of the given series, \( S_n = \Sigma T_n \):
\( S_n = \Sigma \left( \frac{n(n+1)}{2} \right) \)
\( S_n = \frac{1}{2} \Sigma (n^2 + n) \)
\( S_n = \frac{1}{2} (\Sigma n^2 + \Sigma n) \)
Apply the summation formulas:
\( S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \)
Take \( \frac{n(n+1)}{2} \) as a common factor inside the brackets:
\( S_n = \frac{1}{2} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] \)
\( S_n = \frac{n(n+1)}{4} \left[ \frac{2n+1+3}{3} \right] \)
\( S_n = \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] \)
\( S_n = \frac{n(n+1) \cdot 2(n+2)}{4 \cdot 3} \)
\( S_n = \frac{n(n+1)(n+2)}{6} \)
This is the sum of n terms for the first series. Recognizing that each term is a sum itself is key to finding its general form.
(ii) Given series: \( 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... \)
The nth term \( T_n \) of this series is the sum of the squares of the first n natural numbers:
\( T_n = 1^2 + 2^2 + 3^2 + ... + n^2 \)
Using the formula for the sum of squares of the first n natural numbers:
\( T_n = \frac{n(n+1)(2n+1)}{6} \)
Now, find the sum of n terms of the given series, \( S_n = \Sigma T_n \):
\( S_n = \Sigma \left( \frac{n(n+1)(2n+1)}{6} \right) \)
\( S_n = \frac{1}{6} \Sigma (n(n+1)(2n+1)) \)
First, expand the expression inside the summation:
\( n(n+1)(2n+1) = (n^2+n)(2n+1) = 2n^3 + n^2 + 2n^2 + n = 2n^3 + 3n^2 + n \)
So, \( S_n = \frac{1}{6} \Sigma (2n^3 + 3n^2 + n) \)
\( S_n = \frac{1}{6} (2\Sigma n^3 + 3\Sigma n^2 + \Sigma n) \)
Apply the summation formulas:
\( S_n = \frac{1}{6} \left[ 2\left(\frac{n(n+1)}{2}\right)^2 + 3\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \)
\( S_n = \frac{1}{6} \left[ \frac{2n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right] \)
\( S_n = \frac{1}{6} \left[ \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right] \)
Take \( \frac{n(n+1)}{2} \) as a common factor inside the brackets:
\( S_n = \frac{1}{6} \cdot \frac{n(n+1)}{2} \left[ n(n+1) + (2n+1) + 1 \right] \)
\( S_n = \frac{n(n+1)}{12} [n^2 + n + 2n + 1 + 1] \)
\( S_n = \frac{n(n+1)}{12} [n^2 + 3n + 2] \)
Factor the quadratic expression \( n^2 + 3n + 2 \). It factors into \( (n+1)(n+2) \):
\( S_n = \frac{n(n+1)(n+1)(n+2)}{12} \)
\( S_n = \frac{n(n+1)^2(n+2)}{12} \)
This is the sum of n terms for the second series. Summing sums of powers is a common challenge, requiring careful application of formulas.
In simple words: For series where each term is already a sum of natural numbers or their squares, first find a formula for that individual term. Then, use the overall summation formulas again to sum up all those terms.

🎯 Exam Tip: Pay close attention to nested summations. First find the formula for the 'k'th term of the *outer* series (which itself is a sum), and then apply the summation to *that* formula to find the overall sum of 'n' terms.