Question

The value of 14^2 +15^2 + ... + 30^2 = ?​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\: {14}^{2} + {15}^{2} + {16}^{2} + - - - + {30}^{2}$

can be rewritten as

$\rm= {1}^{2} + {2}^{2} + {3}^{2} + - - - + {30}^{2} - [{1}^{2} + {2}^{2} + - - + {13}^{2}]$

$\rm \:  =  \: \displaystyle\sum_{k=1}^{30} {k}^{2} \: - \: \displaystyle\sum_{k=1}^{13} {k}^{2}$

We know,

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{30(30 + 1)(60 + 1)}{6} - \dfrac{13(13 + 1)(26 + 1)}{6}$

$\rm \:  =  \: 5 \times 31 \times 61 - 13 \times 7 \times 9$

$\rm \:  =  \: 9455 - 819$

$\rm \:  =  \: 8636$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ {14}^{2} + {15}^{2} + {16}^{2} + - - - + {30}^{2} = 8636}}$

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$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{3} = \bigg[\frac{n(n + 1)}{2}\bigg] ^{2} \: }}$