Question

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

Answer 1

Answer:

hello :

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

Step-by-step 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

Step-by-step explanation:

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