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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 3 of 160

Marks: +1, -0

1+(1+3)+(1+3+5)+(1+3+5+7)+\dots

=

385
285
506
406

Solution:

Given series

1+(1+3)+(1+3+5)+(1+3+5+7)+\dots

to 10 terms.

The

n

th term of the series is the sum of first

n

odd numbers.

So,

T_n=1+3+5+\dots+(2n-1)=n^2

\therefore\; T_1=1^2=1

T_2=2^2=4

T_3=3^2=9

T_{10}=10^2=100

Thus,

S_{10}=\sum\limits_{n=1}^{10} n^2

We know that the sum of the first

k

squares is

\sum\limits_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}

For

k=10

S_{10}\;=\;\frac{10(10+1)(2 \times 10+1)}{6}

\;=\;\frac{10 \times 11 \times 21}{6}\;=\;\frac{2310}{6}=385

\therefore S_{10}=385

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