Maharashtra Board Class 11 Maths Part 1 Chapter 4 Sequences and Series 4.5 Solutions

Question 1. Find the sum \(\sum_{r=1}^n (r + 1)(2r - 1)\).
Answer:
Solution:
\[ \sum_{r=1}^n (r+1)(2r-1) \] \[ = \sum_{r=1}^n (2r^2+r-1) \] \[ = 2 \sum_{r=1}^n r^2 + \sum_{r=1}^n r - \sum_{r=1}^n 1 \] \[ = 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} - n \] \[ = \frac{n}{6} [2(2n^2 + 3n + 1) + 3(n + 1) - 6] \] \[ = \frac{n}{6} (4n^2 + 6n + 2 + 3n + 3 - 6) \] \[ = \frac{n}{6} (4n^2 + 9n - 1) \] In simple words: To find the sum, first expand the product \((r+1)(2r-1)\), then separate the sum into three standard summation formulas for \(r^2\), \(r\), and a constant, and finally simplify the resulting algebraic expression.

🎯 Exam Tip: Remember the standard summation formulas for \(\sum r\), \(\sum r^2\), and \(\sum 1\) to efficiently solve series problems. Pay attention to algebraic simplification.

Question 2. Find \(\sum_{r=1}^n (3r^2 - 2r + 1)\).
Answer:
Solution:
\[ \sum_{r=1}^n (3r^2 - 2r + 1) \] \[ = 3 \sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r + \sum_{r=1}^n 1 \] \[ = 3 \cdot \frac{n(n+1)(2n+1)}{6} - 2 \frac{n(n+1)}{2} + n \] \[ = \frac{n}{2} [(2n^2 + 3n + 1) - 2(n + 1) + 2] \] \[ = \frac{n}{2} (2n^2 + 3n + 1 - 2n - 2 + 2) \] \[ = \frac{n}{2} (2n^2 + n + 1) \] In simple words: The sum of the given quadratic expression is found by applying the linearity of summation and using the standard formulas for the sum of squares, sum of natural numbers, and sum of a constant, then simplifying the algebraic result.

🎯 Exam Tip: Direct application of summation properties and standard formulas is key here. Ensure careful algebraic expansion and combination of terms for accuracy.

Question 3. Find \(\sum_{r=1}^n \frac{1+2+3+\dots+r}{r}\).
Answer:
Solution:
\[ \sum_{r=1}^n \frac{1+2+3+\dots+r}{r} \] We know that \(1+2+3+\dots+r = \frac{r(r+1)}{2}\). Therefore, \[ = \sum_{r=1}^n \frac{r(r+1)}{2r} \] \[ = \sum_{r=1}^n \frac{(r+1)}{2} \] \[ = \frac{1}{2} \sum_{r=1}^n (r+1) \] \[ = \frac{1}{2} \left[ \sum_{r=1}^n r + \sum_{r=1}^n 1 \right] \] \[ = \frac{1}{2} \left[ \frac{n(n+1)}{2} + n \right] \] \[ = \frac{n}{4} [(n+1) + 2] \] \[ = \frac{n}{4} (n+3) \] In simple words: To find this sum, first simplify the numerator using the formula for the sum of the first \(r\) natural numbers, then cancel out \(r\) terms and apply the standard summation formulas for \(r\) and a constant.

🎯 Exam Tip: Recognize and simplify the arithmetic series in the numerator before applying summation formulas. Algebraic simplification after applying formulas is crucial.

Question 4. Find \(\sum_{r=1}^n \frac{1^3+2^3+3^3+\dots+r^3}{r(r+1)}\).
Answer:
Solution:
We know that,
\[ 1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4} \] Therefore,
\[ 1^3 + 2^3 + 3^3 + \dots + r^3 = \frac{r^2(r+1)^2}{4} \] So, the term inside the summation is:
\[ \frac{1^3 + 2^3 + 3^3 + \dots + r^3}{r(r+1)} = \frac{\frac{r^2(r+1)^2}{4}}{r(r+1)} \] \[ = \frac{r(r+1)}{4} \] Now, find the sum:
\[ \sum_{r=1}^n \frac{r(r+1)}{4} \] \[ = \frac{1}{4} \sum_{r=1}^n (r^2+r) \] \[ = \frac{1}{4} \left[ \sum_{r=1}^n r^2 + \sum_{r=1}^n r \right] \] \[ = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \] \[ = \frac{1}{4} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] \] \[ = \frac{n(n+1)}{8} \left[ \frac{2n+1+3}{3} \right] \] \[ = \frac{n(n+1)}{8} \left[ \frac{2n+4}{3} \right] \] \[ = \frac{n(n+1)(2n+4)}{24} \] \[ = \frac{2n(n+1)(n+2)}{24} \] \[ = \frac{n(n+1)(n+2)}{12} \] In simple words: This problem requires using the formula for the sum of cubes to simplify the numerator, then simplifying the fraction and applying the standard summation formulas for \(r^2\) and \(r\) to find the final sum.

🎯 Exam Tip: Correctly identifying and applying the sum of cubes formula is the first critical step. Algebraic simplification after applying summation formulas must be meticulous to avoid errors.

Question 5. Find the sum \(5 \times 7 + 7 \times 9 + 9 \times 11 + 11 \times 13 + \dots\) upto n terms.
Answer:
Solution:
The given series is \(5 \times 7 + 7 \times 9 + 9 \times 11 + 11 \times 13 + \dots\) upto n terms
Now, consider the first terms of each product: \(5, 7, 9, 11, \dots\) are in A.P.
Its rth term \(= 5 + (r - 1)(2) = 5 + 2r - 2 = 2r + 3\)
Consider the second terms of each product: \(7, 9, 11, \dots\) are in A.P.
Its rth term \(= 7 + (r - 1)(2) = 7 + 2r - 2 = 2r + 5\)
So, the rth term of the given series is \((2r + 3)(2r + 5)\).
The sum S to n terms is:
\[ S = \sum_{r=1}^n (2r+3)(2r+5) \] \[ = \sum_{r=1}^n (4r^2 + 10r + 6r + 15) \] \[ = \sum_{r=1}^n (4r^2 + 16r + 15) \] \[ = 4 \sum_{r=1}^n r^2 + 16 \sum_{r=1}^n r + 15 \sum_{r=1}^n 1 \] \[ = 4 \frac{n(n+1)(2n+1)}{6} + 16 \frac{n(n+1)}{2} + 15n \] \[ = \frac{2n(n+1)(2n+1)}{3} + 8n(n+1) + 15n \] \[ = \frac{n}{3} [2(n+1)(2n+1) + 24(n+1) + 45] \] \[ = \frac{n}{3} [2(2n^2 + 3n + 1) + 24n + 24 + 45] \] \[ = \frac{n}{3} [4n^2 + 6n + 2 + 24n + 69] \] \[ = \frac{n}{3} (4n^2 + 30n + 71) \] In simple words: First, find the general rth term of the series by identifying the arithmetic progression of the factors in each product. Once the rth term is found, expand it, and then use standard summation formulas for \(r^2\), \(r\), and a constant to calculate the sum.

🎯 Exam Tip: Identifying the general term (rth term) of a series is often the most critical step. Break down complex products into simpler sums using standard formulas and simplify algebraic expressions meticulously.

Question 6. Find the sum \(2^2 + 4^2 + 6^2 + 8^2 + \dots\) upto n terms.
Answer:
Solution:
The given series is \(2^2 + 4^2 + 6^2 + 8^2 + \dots\) upto n terms.
This can be written as:
\[ (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + (2 \times 4)^2 + \dots \] The rth term is \((2r)^2\).
So, the sum S to n terms is:
\[ S = \sum_{r=1}^n (2r)^2 \] \[ = \sum_{r=1}^n 4r^2 \] \[ = 4 \sum_{r=1}^n r^2 \] \[ = 4 \frac{n(n+1)(2n+1)}{6} \] \[ = \frac{2n(n+1)(2n+1)}{3} \] In simple words: To find the sum, express each term as a multiple of a natural number squared, factor out the common constant, and then apply the standard summation formula for the sum of squares.

🎯 Exam Tip: Recognize patterns in the series to express the rth term simply. Factoring out constants before applying summation formulas can simplify calculations.

Question 7. Find \((70^2 - 69^2) + (68^2 - 67^2) + (66^2 - 65^2) + \dots + (2^2 - 1^2)\).
Answer:
Solution:
Let S = \((70^2 - 69^2) + (68^2 - 67^2) + \dots + (2^2 - 1^2)\)
Rearranging in ascending order:
S = \((2^2 - 1^2) + (4^2 - 3^2) + \dots + (70^2 - 69^2)\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\):
S = \((2-1)(2+1) + (4-3)(4+3) + \dots + (70-69)(70+69)\)
S = \(1(2+1) + 1(4+3) + \dots + 1(70+69)\)
S = \((2+1) + (4+3) + \dots + (70+69)\)
S = \(3 + 7 + 11 + \dots + 139\)
This is an Arithmetic Progression (A.P.) with first term \(a = 3\) and common difference \(d = 4\).
The general term of this A.P. is \(T_r = a + (r-1)d = 3 + (r-1)4 = 3 + 4r - 4 = 4r - 1\).
To find the number of terms (n), set the last term equal to \(T_r\):
\(139 = 4n - 1\)
\(140 = 4n\)
\(n = 35\)
Now, we need to find the sum of this A.P. S = \(\frac{n}{2}(a + l)\) where \(l\) is the last term.
S = \(\frac{35}{2}(3 + 139)\)
S = \(\frac{35}{2}(142)\)
S = \(35 \times 71\)
S = \(2485\)

Alternatively, from \(S = \sum_{r=1}^{35} (4r-1)\):
\[ S = 4 \sum_{r=1}^{35} r - \sum_{r=1}^{35} 1 \] \[ S = 4 \frac{35(35+1)}{2} - 35 \] \[ S = 4 \frac{35 \times 36}{2} - 35 \] \[ S = 4 \times 35 \times 18 - 35 \] \[ S = 140 \times 18 - 35 \] \[ S = 2520 - 35 \] \[ S = 2485 \] In simple words: Simplify each term using the difference of squares formula, which turns the expression into a sum of an arithmetic progression. Then, find the number of terms and calculate the sum of the arithmetic progression.

🎯 Exam Tip: Recognizing the difference of squares identity \((a^2 - b^2)\) is critical. Convert the series into a standard arithmetic progression and use its sum formula for efficiency.

Question 8. Find the sum \(1 \times 3 \times 5 + 3 \times 5 \times 7 + 5 \times 7 \times 9 + \dots + (2n - 1)(2n + 1)(2n + 3)\).
Answer:
Solution:
The given series is \(1 \times 3 \times 5 + 3 \times 5 \times 7 + 5 \times 7 \times 9 + \dots + (2n - 1)(2n + 1)(2n + 3)\)
Now, let's find the rth term.
For the first factor: \(1, 3, 5, \dots\) are in A.P. with \(a = 1\) and \(d = 2\).
The rth term \(= 1 + (r - 1)2 = 1 + 2r - 2 = 2r - 1\).
For the second factor: \(3, 5, 7, \dots\) are in A.P. with \(a = 3\) and \(d = 2\).
The rth term \(= 3 + (r - 1)2 = 3 + 2r - 2 = 2r + 1\).
For the third factor: \(5, 7, 9, \dots\) are in A.P. with \(a = 5\) and \(d = 2\).
The rth term \(= 5 + (r - 1)2 = 5 + 2r - 2 = 2r + 3\).
So, the rth term of the series is \(T_r = (2r - 1)(2r + 1)(2r + 3)\).
The sum S to n terms is:
\[ S = \sum_{r=1}^n (2r - 1)(2r + 1)(2r + 3) \] \[ = \sum_{r=1}^n (4r^2 - 1)(2r + 3) \] \[ = \sum_{r=1}^n (8r^3 + 12r^2 - 2r - 3) \] \[ = 8 \sum_{r=1}^n r^3 + 12 \sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r - 3 \sum_{r=1}^n 1 \] \[ = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \frac{n(n+1)(2n+1)}{6} - 2 \frac{n(n+1)}{2} - 3n \] \[ = 8 \frac{n^2(n+1)^2}{4} + 2n(n+1)(2n+1) - n(n+1) - 3n \] \[ = 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n \] Now, factor out n:
\[ = n [2n(n+1)^2 + 2(n+1)(2n+1) - (n+1) - 3] \] \[ = n [(n+1)(2n(n+1) + 2(2n+1) - 1) - 3] \] \[ = n [(n+1)(2n^2 + 2n + 4n + 2 - 1) - 3] \] \[ = n [(n+1)(2n^2 + 6n + 1) - 3] \] \[ = n [2n^3 + 6n^2 + n + 2n^2 + 6n + 1 - 3] \] \[ = n [2n^3 + 8n^2 + 7n - 2] \] In simple words: Determine the general rth term by finding the pattern in each factor of the products. Expand the cubic expression for the rth term, then apply the standard summation formulas for \(r^3\), \(r^2\), \(r\), and a constant, followed by algebraic simplification.

🎯 Exam Tip: Finding the rth term is critical. Be extremely careful with algebraic expansion of cubic terms and subsequent simplification of fractions involving \(n\). Using square brackets for intermediate steps helps manage complexity.

Question 9. Find n, if \(\frac{1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \dots + \text{upto n terms}}{1+2+3+4+\dots+\text{upto n terms}} = \frac{100}{3}\).
Answer:
Solution:
The given equation is:
\[ \frac{1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + \text{upto n terms}}{1+2+3+4+\dots+\text{upto n terms}} = \frac{100}{3} \] Let's find the numerator sum (N) and denominator sum (D).
Numerator: \(N = 1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + n(n+1)\)
The rth term is \(r(r+1) = r^2 + r\).
\[ N = \sum_{r=1}^n (r^2 + r) = \sum_{r=1}^n r^2 + \sum_{r=1}^n r \] \[ N = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \] \[ N = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] \] \[ N = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} \right] \] \[ N = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} \right] \] \[ N = \frac{n(n+1)2(n+2)}{6} \] \[ N = \frac{n(n+1)(n+2)}{3} \] Denominator: \(D = 1+2+3+4+\dots+n\)
\[ D = \sum_{r=1}^n r = \frac{n(n+1)}{2} \] Now substitute N and D back into the equation:
\[ \frac{\frac{n(n+1)(n+2)}{3}}{\frac{n(n+1)}{2}} = \frac{100}{3} \] \[ \frac{n(n+1)(n+2)}{3} \cdot \frac{2}{n(n+1)} = \frac{100}{3} \] Cancel out common terms \(n(n+1)\) (assuming \(n \neq 0, n \neq -1\), which is true for a sum of terms):
\[ \frac{2(n+2)}{3} = \frac{100}{3} \] Multiply both sides by 3:
\[ 2(n+2) = 100 \] Divide by 2:
\[ n+2 = 50 \] \[ n = 48 \] In simple words: First, calculate the sum of the series in the numerator and the sum of the series in the denominator using standard summation formulas. Then, substitute these sums into the given equation and solve for \(n\) algebraically.

🎯 Exam Tip: Correctly identifying the rth term for both the numerator and denominator series is vital. Efficiently applying standard sum formulas and simplifying the resulting algebraic equation is crucial for an accurate solution.

Question 10. If S1, S2, and S3 are the sums of first n natural numbers, their squares, and their cubes respectively, then show that: \(9S_2^2 = S_3(1 + 8S_1)\).
Answer:
Solution:
Given:
\(S_1 = 1+2+3+\dots+n = \sum_{r=1}^n r = \frac{n(n+1)}{2}\)
\(S_2 = 1^2+2^2+3^2+\dots+n^2 = \sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}\)
\(S_3 = 1^3+2^3+3^3+\dots+n^3 = \sum_{r=1}^n r^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}\)

We need to show: \(9S_2^2 = S_3(1 + 8S_1)\).

Let's evaluate the Left Hand Side (L.H.S.):
\[ \text{L.H.S.} = 9S_2^2 \] \[ = 9 \left[ \frac{n(n+1)(2n+1)}{6} \right]^2 \] \[ = 9 \frac{n^2(n+1)^2(2n+1)^2}{36} \] \[ = \frac{n^2(n+1)^2(2n+1)^2}{4} \] Now, let's evaluate the Right Hand Side (R.H.S.):
\[ \text{R.H.S.} = S_3(1 + 8S_1) \] \[ = \frac{n^2(n+1)^2}{4} \left( 1 + 8 \cdot \frac{n(n+1)}{2} \right) \] \[ = \frac{n^2(n+1)^2}{4} \left( 1 + 4n(n+1) \right) \] \[ = \frac{n^2(n+1)^2}{4} \left( 1 + 4n^2 + 4n \right) \] \[ = \frac{n^2(n+1)^2}{4} (2n+1)^2 \] Since L.H.S. = R.H.S., the identity is proved.
\[ 9S_2^2 = S_3(1 + 8S_1) \] In simple words: To prove the identity, substitute the standard formulas for \(S_1\), \(S_2\), and \(S_3\) into both sides of the equation. Simplify both the Left Hand Side and the Right Hand Side separately to show that they are equal.

🎯 Exam Tip: Accurately writing down the formulas for \(S_1, S_2, S_3\) is the foundation. Algebraic manipulation, especially squaring and expanding terms, must be precise to arrive at the correct identity.