Observe the following pattern
$ 1^{2}=\frac{1}{6}[1 \times(1+1) \times(2 \times 1)+1)] $
$ 1^{2}+2^{2}=\frac{1}{6}[2 \times(2+1) \times(2 \times 2)+1)] $
$ 1^{2}+2^{2}+3^{2}=\frac{1}{6}[3 \times(3+1) \times(2 \times 3)+1)] $
$ 1^{2}+2^{2}+3^{2}+4^{2}=\frac{1}{6}[4 \times(4+1) \times(2 \times 4)+1)] $
and find the values of each of the following:
(i) $1^2 + 2^2 + 3^2 + 4^2 +…………… + 10^2$
(ii)$5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2$

To do:

We have to find the values of the given series.

Solution:

We observe that,

\( 1^{2}=\frac{1}{6}[1 \times(1+1) \times(2 \times 1)+1)] \)

\( 1^{2}+2^{2}=\frac{1}{6}[2 \times(2+1) \times(2 \times 2)+1)] \)

\( 1^{2}+2^{2}+3^{2}=\frac{1}{6}[3 \times(3+1) \times(2 \times 3)+1)] \)
\( 1^{2}+2^{2}+3^{2}+4^{2}=\frac{1}{6}[4 \times(4+1) \times(2 \times 4)+1)] \)

Therefore,

(i) $1^{2}+2^{2}+3^{2}+4^{2}+\ldots .+10^{2}=\frac{1}{6}\{10 \times(10+1) \times(2 \times 10+1)]$

$=\frac{1}{6}[10 \times 11 \times 21]$

$=\frac{10 \times 11 \times 21}{6}$

$=\frac{2310}{6}$

$=385$

(ii) $5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}=[1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+12^{2}]-[1^{2}+2^{2}+3^{2}+4^{2}]$

$=\frac{1}{6}[12 \times(12+1) \times(2 \times 12+1)]-\frac{1}{6}[4 \times(4+1) \times(2 \times 4+1)]$

$=\frac{1}{6}[12 \times13 \times25]-\frac{1}{6}[4 \times5\times9]$

$=650-30$

$=620$