Question

9. sum of the 66 consecutive integers is 5181, then sum of the squares is?

Answer 1

let the consecutives numbers are x,x+1,x+2 ... x+65

sum of these numbers are x+x+1+x+2...x+65 = 5181

66x+ (1+2+3+...+65) = 5181

it is an AP 1+2+3+4...65

a =1 ,d = 1, l= 65

S = n/2 (a+l)

S = 65/2(1+65) =   2145

66x +2145 = 5181

66 x = 5181-2145

x = 3036/66 = 46

So,the numbers are 46,47,48,49,50 ...111

Now calculate the sum of their squares are x² + (x+1)²+(x+2)²...+(x+65)²

= x²+x²+2x+1+x²+4x+4+x²+6x+9...x²+130x+4225

= 66 x²+ x ( 2 +4 +6 + ...+130) +( 1+4+9+16 ...+4225)

= 66 x² + x [( 65/2)(2+130)]+ (1² + 2²+3²+4²...+65²)

(1² + 2²+3²+4²...+65²): sum of this series is  [n(n+1)(2n+1)/6]

= 66x²+ 4290 x + [n(n+1)(2n+1)/6]

= 66 (46)²+ 4290×46 +[65(65+1)(130+1)/6]

= 66×2116 + 197,340 + 93,665

= 139,656+197,340 + 93,665

= 430,661