Homework Sourse716 Show that 1222 n2nn12n16 Solution Let fn nn12n16 Then
7.16 Show that 1^2+2^2+. . .+n^2=n(n+1)(2n+1)/6. ![7.16 Show that 1^2+2^2+. . .+n^2=n(n+1)(2n+1)/6. Solution Let f(n) = n(n+1)(2n+1)/6 Then f(n) - f(n-1) = n(n+1)(2n+1)/6 - (n-1)[(n-1)+1][2(n-1)+1]/6, = n(n+1)(](/WebImages/14/716-show-that-1222-n2nn12n16-solution-let-fn-nn12n16-then-1017723-1761526036-0.webp "7.16 Show that 1^2+2^2+. . .+n^2=n(n+1)(2n+1)/6. Solution Let f(n) = n(n+1)(2n+1)/6 Then f(n) - f(n-1) = n(n+1)(2n+1)/6 - (n-1)[(n-1)+1][2(n-1)+1]/6, = n(n+1)(")
Solution
Let f(n) = n(n+1)(2n+1)/6 Then f(n) - f(n-1) = n(n+1)(2n+1)/6 - (n-1)[(n-1)+1][2(n-1)+1]/6, = n(n+1)(2n+1)/6 - (n-1)n(2n-1)/6, = n[(2n^2+3n+1)-(2n^2-3n+1)]/6, = n(6n)/6, = n2. Therefore 12 = f(1) - f(0), 22 = f(2) - f(1), 32 = f(3) - f(2), ... ... (n-1)2 = f(n-1) - f(n-2), n2 = f(n) - f(n-1). Now if we a dd all these, and see that on the right side, then there will be cancellation towards right side wich we will get 1^2 + 2^2 + 3^2 + ... + n^2 = f(n) - f(0) = n(n+1)(2n+1)/6.