A nice sum

sum of squares from 1 to n = n * (n+1) * (2n+1) / 6

here n =8

by substituting n = 8 we get 8 * (8+1 ) * (16+1) / 6

= 8 * 9 * 17 / 6

= 204 :)

=1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2 =1+4+9+16+25+36+49+64=204 AND FORMULA Sum= [N(N+1)(2N+1)]/6 N=8 =(8x9x17)/6=4x3x17=204

n(n+1)(2n+1)/6 where n=8

Here, Lower limit of n=1 and Upper limit of n=8 Then,

1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2 204

sum of squares from 1 to n = n * (n+1) * (2n+1) / 6 here n =8 by substituting n = 8 we get 8 * (8+1 ) * (16+1) / 6 = 8 * 9 * 17 / 6 = 204

n(n+1)(2n+1)/6, sum of squares up to n.

Plugging in 8, we get

8(9)(17)/6=204.

Its better way to use the formula. That is -

1^2 + 2^2 + 3^2 + . . . . . . + n^2 = {n * (n+1) * (2n+1)}/6 ; Where n=8

Took square of each digit from 1 to 8 and added them up.

Ans : 204 sum of squares of 1 to 8 is the answer

that's so easy SUM= n * (n+1) * (2n+1) / 6.

In this question have three question mark ,and options are given in single,two,three,and four digit numbers ,here 204 is only three digit number so i choose that correct one taht's it.

Well My simple logic by seeing multiple choices is 8x8=64 So it's impossible for 12 and 9 Then if 8x8=64, below of 8 (7x7, 6x6), they won't be more than 64 100x8 = 800 < 5463 So it's really impossible too 204 must be true hehe

1+4+9+16+25+36+49+64=204

(1×1)+(2×2)+(3×3)+(4×4)+(5×5)+(6×6)+(7×7)+(8×8) =1+4+9+16+25+36+49+64=204

I did the same as Jithin, wihout any formula, 8² = 64, so... if 64 > 12 & 9, this options were dropped, and only I have 2 options, 5463 & 204, so the ans is 204

1x1+2x2+3x3+4x4+5x5+6x6+7x7+8x8=? ; then
1+4+9+16+25+36+49+64= 204