Hence P(n) is true for all natural numbers n.ExampleFor all n 2 1, pro

1.	Hence P(n) is true for all natural numbers n.ExampleFor all n 2 1, prove that1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
Answer» Proof: I. Prove that the equation holds for n = 1 If n = 1, then [n(n+1)(2n+1)]/6 = [(1)(2)(3)]/6 = 1. So,12= [(1)(1+1)(2(1)+1)]/6 II. Assume that the equation holds for n and prove that the equation holds for n+1. Assume: 12+22+ ... + n2= [n(n+1)(2n+1)]/6 Prove: 12+ 22+ ... + n2+(n+1)2= [(n+1)(n+2)(2n+3)]/6 By assumption, 12+ 22+... + n2+ (n+1)2= [n(n+1)(2n+1)]/6 + (n+1)2 = [n(n+1)(2n+1) +6(n+1)2]/6 = [(n+1){n(2n+1) + 6(n+1)}]/6 =[(n+1){2n2+ 7n +6}]/6 = [(n+1)(n+2)(2n+3)]/6 Therefore, by induction on n, the equation is valid for all positive integers, n.

Hence P(n) is true for all natural numbers n.ExampleFor all n 2 1, prove that1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}

Answer»

Proof:

I. Prove that the equation holds for n = 1

If n = 1, then [n(n+1)(2n+1)]/6 = [(1)(2)(3)]/6 = 1.

So,12= [(1)(1+1)(2(1)+1)]/6

II. Assume that the equation holds for n and prove that the equation holds for n+1.

Assume: 12+22+ ... + n2= [n(n+1)(2n+1)]/6

Prove: 12+ 22+ ... + n2+(n+1)2=

[(n+1)(n+2)(2n+3)]/6

By assumption, 12+ 22+... + n2+ (n+1)2=

[n(n+1)(2n+1)]/6 + (n+1)2

= [n(n+1)(2n+1) +6(n+1)2]/6

= [(n+1){n(2n+1) + 6(n+1)}]/6

=[(n+1){2n2+ 7n +6}]/6

= [(n+1)(n+2)(2n+3)]/6

Therefore, by induction on n, the equation is valid for all positive integers, n.