Question

Find the sum to ${}^{\prime }{n}^{\prime }$ terms of the series

$5^2+6^2+7^2+.........+{20}^2$

Answer 1

The given series is $5^2+6^2+7^2+.........+{20}^2$

We can write the given series as $(1^2+2^2+3^2+4^2+5^2+6^2+7^2+.........+{20}^2)-(1^2+2^2+3^2+4^2)$

We know that square of the ${}^{\prime }{n}^{\prime }$ natural number is given by $i.e,(1^2+2^2+3^2.....+n^2)$

$=\frac{n(n+1)(2n+1)}{6}\dots \dots \dots \left(i\right)$

On putting $n=20$ in the above $eq.\left(i\right),$ we get

$(1^2+2^2+3^2+4^2+5^2+6^2+7^2+.........+{20}^2)=\frac{20(20+1)(2\times 20+1)}{6}$

$=\frac{20\left(21\right)(40+1)}{6}$

$=\frac{20\left(21\right)\left(41\right)}{6}=\frac{20\left(7\right)\left(41\right)}{2}=\frac{10\left(7\right)\left(41\right)}{1}=2870$

Now, on putting $n=4$ in the $eq.\left(i\right)$, we get

$\therefore (1^2+2^2+3^2+4^2)=\frac{4(4+1)(2\times 4+1)}{6}$

$=\frac{4\left(5\right)(8+1)}{6}=\frac{4\left(5\right)\left(9\right)}{6}=\frac{4\left(5\right)\left(3\right)}{2}=30$

So, $5^2+6^2+7^2+.........+{20}^2=2870-30=2840$ [Using above values]

Thus, the sum of ${}^{\prime }{n}^{\prime }$ terms of series $5^2+6^2+7^2+.......+{20}^2$ is $2840.$