SATHEE: Sequences and Series 5 Question 21

Sequences and Series 5 Question 21

12. If $S_{1}, S_{2}, S_{3}, \dots, S_{n}$ are the sums of infinite geometric series, whose first terms are $1, 2, 3, \dots, n$ and whose common ratios are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n + 1}$ respectively, then find the values of $S_{1}^{2} + S_{2}^{2} + S_{3}^{2} + \dots + S_{2 n - 1}^{2}$ .

(1991, 4M)

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Solution:

PLAN Write the $n$ th term of the given series and simplify it to get its lowest form. Then, apply, $S_{n} = \sum T_{n}$

Given series is $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots$

Let $T_{n}$ be the $n$ th term of the given series.

$\begin{aligned} ∴ T_{n} & = \frac{1^{3} + 2^{3} + 3^{3} + \dots + n^{3}}{1 + 3 + 5 + \dots + upto n terms} \\ = \frac{\frac{n (n + 1)^{2}}{2}}{n^{2}} = \frac{(n + 1)^{2}}{4} \\ S_{9} & = \sum_{n = 1}^{9} \frac{(n + 1)^{2}}{4} = \frac{1}{4} (2^{2} + 3^{2} + \dots + 10^{2}) + 1^{2} - 1^{2}] \\ = \frac{1}{4} \frac{10 (10 + 1) (20 + 1)}{6} - 1 = \frac{384}{4} = 96 \end{aligned}$

Sequences and Series 5 Question 21

Solution:

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Sequences and Series 5 Question 21

12. If S1,S2,S3,…,Sn are the sums of infinite geometric series, whose first terms are 1,2,3,…,n and whose common ratios are 12,13,14,…,1n+1 respectively, then find the values of S12+S22+S32+…+S2n−12.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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