Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$$
5^{2}+6^{2}+7^{2}+\ldots \ldots \ldots \ldots \ldots+20^{2}=
$$
MathematicsSequences and SeriesMHT CETMHT CET 2020 (12 Oct Shift 2)
Options:
  • A 2860
  • B 2840
  • C 2830
  • D 2850
Solution:
1167 Upvotes Verified Answer
The correct answer is: 2840
The given series is $5^{2}+6^{2}+7^{2}+\ldots+20^{2} n^{\text {th }}$ term,
$\begin{aligned}
a_{n} &=(n+4)^{2}=n^{2}+8 n+16 \\
\therefore S_{n} &=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(k^{2}+8 k+16\right) \\
&=\sum_{k=1}^{n} k^{2}+8 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 16 \\
&=\frac{n(n+1)(2 n+1)}{6}+\frac{8 n(n+1)}{2}+16 n
\end{aligned}$
$16^{\text {th }}$ term is $(16+4)^{2}=20^{2}$
$\begin{aligned}
\therefore S_{10} &=\frac{16(16+1)(2 \times 16+1)}{6}+\frac{8 \times 16 \times(16+1)}{2}+16 \times 16 \\
&=\frac{(16)(17)(33)}{6}+\frac{(8) \times 16 \times(16+1)}{2}+16 \times 16 \\
&=\frac{(16)(17)(33)}{6}+\frac{(8)(16)(17)}{2}+256 \\
&=1496+1088+256 \\
&=2840 \\
\therefore 5^{2} &+6^{2}+7^{2}+\ldots \ldots+20^{2}=2840
\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.