Find the sum of squares of all numbers between 10 and 70 and which are divisible by 4.
Hello, rickrishav!
I hope I interpreted the problem correctly . . .
Find the sum of squares of all integers between 10 and 70 which are divisible by 4.
The integers between 10 and 70 which are multiple of 4 are:
. . $\displaystyle \{12,16,20,24,\,\hdots\,68\} \;=\;\{3\!\cdot\!4,\:4\!\cdot\!4,\:5\!\cdot\!4,\:6 \!\cdot\!6,\:\hdots\:17\!\cdot\!4\}$
The sum of their squares is:
. . $\displaystyle S \;=\;3^2\!\cdot\!4^2 + 4^2\!\cdot\!4^2 + 5^2\!\cdot\!4^2 + 5^2\!\cdot4^2 + \hdots + 17^2\!\cdot\!4^2$
. . . . $\displaystyle =\;4^2\left(3^2 + 4^2 + 5^2 + \hdots + 17^2\right)$
. . . . $\displaystyle \displaystyle = \;4^2\left(\sum^{17}_{k=1} k^2 - \sum^2_{k=1}k^2\right) $
. . . . $\displaystyle -\; 4^2\left(\dfrac{17\!\cdot\!18\!\cdot\!35}{6} - \dfrac{2\!\cdot\!3\!\cdot\!5}{6}\right) $ .**
. . . . $\displaystyle =\;16\cdot 1780$
. . . . $\displaystyle =\;28,\!480$
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The sum of the first $\displaystyle \,n$ squares is:
. . $\displaystyle \displaystyle \sum^n_{k+1} k^2 \;=\; 1^2 + 2^2 + 3^2 + \hdots + n^2 \;=\;\frac{n(n+1)(2n+1)}{6}$