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Questions  >>  CBSE XI  >>  Math  >>  Sequences and Series

Find the sum of 20 terms of the series. $5^2+6^2+7^2+.....20^2$

1 Answer

  • $n^{th}$ term of an A.P$=a+(n-1)d$
  • $\sum n=\large\frac{n(n+1)}{2}$
  • $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
  • $\sum k=nk$ where $k$ is a constant.
Given series is $5^2+6^2+7^2+........20^2$
We have to find the $n^{th}$ term of this series.
Each term in this series is square of a number.
$i.e.,$ The $n^{th}$ term of the series is square of the $n^{th}$ term
of the sequence $5,6,7...........$
This sequence is an A.P. whose $n^{th}$ term is $5+(n-1).1$
Since, the $n^{th}$ term of an $A.P.$ is $a+(n-1)d$
$\Rightarrow\:4+n=20$ or $n=16$
$i.e., $ The number of terms in the given series is 1$n=16$
$\therefore$ The $n^{th}$ term of the given series is $t_n=(4+n)^2$
Step 2
We have to find the sum of 16 terms of the given series =$S_{16}$
The sum of $n$ terms of the any series is $S_n=\sum \ t_n$
$=\sum (16+8n+n^2)$
$=\sum 16+8\sum n+\sum n^2$
We know that $\sum n=\large\frac{n(n+1)}{2}$ and
$\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
Also we know that $\sum k=kn$
$\Rightarrow \sum 16=16n$
$8\sum n=8\times \large\frac{n(n+1)}{2}$$=4n(n+1)$
$\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
But since we have to find the sum of 16 terms only.
$\therefore n=16$
Substituting the value of $n$ we get
$\sum 16=16\times 16=256$
$8\sum n=4\times 16\times (16+1)=64\times 17=1088$
$\sum n^2=\large\frac{16\times (16+1)(2\times 16+1)}{6}$
$=\large\frac{16\times 17\times 33}{6}$$=1496$
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