Question

Find the sum of the G.P,16^2+17^2+18^2+...+25^2

Answer 1

hello :

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

Explanation:

let : S = 1²+2²+3²+......+25²

A = 16²+17²+18²+...+25²

B = 1²+2²+3²+......+15²

S = (1²+2²+3²+....+15²) + (16²+17²+18²+...+25²)

calculate : S for : n = 25

S = 25(26)(27)/6 = 2925

calculate : B for : n = 15

B = 15(16)(17)/6 = 680

so : S = B + A

A = S - B = 2925 - 680 = 2245

Answer 2

Answer: 4285

Explanation:

`\sum^(25)_(16)i^2=\sum^(25)_(0)i^2-\sum^(15)_(0)i^2\\\\\text{Use the summation formula:}\ (n)/(6)(n+1)(2n+1)\\\\\sum^(25)_(0)i^2 = (25)/(6)(25+1)[2(25)+1]\\\\\\.\qquad=(25(26)(51))/(6)\\\\.\qquad=5525\\\\\\\sum^(15)_(0)i^2 = (15)/(6)(15+1)[2(15)+1]\\\\\\.\qquad=(15(16)(31))/(6)\\\\.\qquad=1240\\\\\\\sum^(25)_(0)i^2-\sum^(15)_(0)i^2\\\\=5525-1240\\\\=4285`