Question

Find 31² + 32² + 33² + .... +50²

Answer 1

it is given that, S = 31² + 32² + 33² + .... + 50²

we know, formula,

1² + 2² + 3² + 4² ..... + n² = n(n + 1)(2n + 1)/6

so here first find $S_{50}$ and then $S_{30}$ and subtract them to get value of S.

i.e., $S=S_{50}-S_{30}$

now, $S_{50}$ = 50(50 + 1)(2 × 50+1)/6 = 50 × 51 × 101/6

similarly, $S_{30}$ = 30 × 31 × 61/6

now, S = 50 × 51 × 101/6 - 30 × 31 × 61/6

= 10/6 [ 5 × 51 × 101 - 3 × 31 × 61 ]

= 10/6 × 20082

= 10 × 3347

= 33470

hence, value of 31² + 32² + 33² + .... + 50² = 33470.

Answer 2

Answer:

Hence the result is=33470

Step-by-step explanation:

We have to find:

$S=31^2+32^2+33^2+.............+50^2$

Formula for sum of square of  natural numbers upto n terms is following:

$S_n=1^2+2^2+3^2+................n^2=\dfrac{(n)(n+1)(2n+1)}{6}$

$S=S_{50}-S_{30}$

$S=1^2+2^2+3^2+................+50^2-1^2+2^2+3^+............+30^2$

$S=\dfrac{(50)(50+1)(2\times50+1)}{6}-\dfrac{(30)(30+1)(2\times30+1)}{6}$

$S=\dfrac{(50)(51)(101)}{6}-\dfrac{(30)(31)(61)}{6}$

After calculating the above term, we get:

$S=\dfrac{257550}{6}-\dfrac{56730}{6}$

After taking LCM 6, We get:

$S=\dfrac{257550-56730}{6}$

After solving the above terms ,we get

$S=\dfrac{200820}{6}$

After dividing ,we get:

S=33470

This is the final result.