$5^{2}+6^{2}+7^{2}+\ldots \ldots \ldots \ldots \ldots+20^{2}=$

$5^{2}+6^{2}+7^{2}+\ldots \ldots \ldots \ldots \ldots+20^{2}=$
  1. 2860
  2. 2840
  3. 2830
  4. 2850

Solution

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 $(16+4)^{2}=20^{2}$ $\begin{aligned} \therefore S_{10} &=\frac{16(16+1)(2 \times 16+1)}{6}+\frac{8 \times 16 \times(16+1)}{2}+16 \times 16 \\ &=\frac{(16)(17)(33)}{6}+\frac{(8) \times 16 \times(16+1)}{2}+16 \times 16 \\ &=\frac{(16)(17)(33)}{6}+\frac{(8)(16)(17)}{2}+256 \\ &=1496+1088+256 \\ &=2840 \\ \therefore 5^{2} &+6^{2}+7^{2}+\ldots \ldots+20^{2}=2840 \end{aligned}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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