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

• Nov 3rd 2010, 07:26 PM
rickrishav
sum of squares of all numbers between 10 and 70 and which are divisible by 4.
Find the sum of squares of all numbers between 10 and 70 and which are divisible by 4.
• Nov 4th 2010, 08:19 AM
Wilmer
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?
• Nov 4th 2010, 10:13 AM
Soroban
Hello, rickrishav!

I hope I interpreted the problem correctly . . .

Quote:

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:

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