sum of squares of all numbers between 10 and 70 and which are divisible by 4.

**rickrishav** · November 3rd 2010, 07:26 PM

Find the sum of squares of all numbers between 10 and 70 and which are divisible by 4.

**Wilmer** · November 4th 2010, 08:19 AM

All evens are divisible by 4; so "sum of even squares":

n = 70/2 = 35; f = (10-2)/2 = 4

2n(n+1)(2n+1)/3 - 2f(f+1)(2f+1)/3 = 59640 - 120 = 59520

"See" that?

**Soroban** · November 4th 2010, 10:13 AM

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:

. . $\{12,16,20,24,\,\hdots\,68\} \;=\;\{3\!\cdot\!4,\:4\!\cdot\!4,\:5\!\cdot\!4,\:6 \!\cdot\!6,\:\hdots\:17\!\cdot\!4\}$

The sum of their squares is:

. . $S \;=\;3^2\!\cdot\!4^2 + 4^2\!\cdot\!4^2 + 5^2\!\cdot\!4^2 + 5^2\!\cdot4^2 + \hdots + 17^2\!\cdot\!4^2$

. . . . $=\;4^2\left(3^2 + 4^2 + 5^2 + \hdots + 17^2\right)$

. . . . $\displaystyle = \;4^2\left(\sum^{17}_{k=1} k^2 - \sum^2_{k=1}k^2\right)$

. . . . $-\; 4^2\left(\dfrac{17\!\cdot\!18\!\cdot\!35}{6} - \dfrac{2\!\cdot\!3\!\cdot\!5}{6}\right)$ .**

. . . . $=\;16\cdot 1780$

. . . . $=\;28,\!480$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

The sum of the first $\,n$ squares is:

. . $\displaystyle \sum^n_{k+1} k^2 \;=\; 1^2 + 2^2 + 3^2 + \hdots + n^2 \;=\;\frac{n(n+1)(2n+1)}{6}$

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