(C) We need to calculate the sum $S = 11^2 + 12^2 + 13^2 + \dots + 20^2$.This can be expressed as the difference of two sums of squares:$S = \sum_{n=1

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(C) We need to calculate the sum $S = 11^2 + 12^2 + 13^2 + \dots + 20^2$.This can be expressed as the difference of two sums of squares:$S = \sum_{n=1}^{20} n^2 - \sum_{n=1}^{10} n^2$.Using the formula $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$:For $k=20$: $\frac{20(21)(41)}{6} = 10 \times 7 \times 41 = 2870$.For $k=10$: $\frac{10(11)(21)}{6} = 5 \times 11 \times 7 = 385$.Therefore,$S = 2870 - 385 = 2485$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Medium

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

A
$2481$
B
$2483$
C
$2485$
D
$2487$

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Class 11

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Sequences and Series

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nth term of special series, Sum to n terms and Infinite number of terms

What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

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If $S_{1}, S_{2}, S_{3}$ are the sum of first $n$ natural numbers,their squares,and their cubes,respectively,show that $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

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If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots + \infty = \frac{\pi^4}{90}$,then the value of $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots + \infty$ is

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If the sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even,what is the sum when $n$ is odd?

Medium

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The sum of the series $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$ is

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$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

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What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

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