Prove by the principle of mathematical induction that for all `n in N

1.	Prove by the principle of mathematical induction that for all `n in N :``1^2+2^2+3^2++n^2=1/6n(n+1)(2n+1)`
Answer» Let the given statement be P(n). Then, `P(n): 1^(2)+2^(2)+3^(2)+…+n^(2)= 1/6 n(n+1)(2n+1)`. Putting n = 1 in the given statement, we get LHS = ` 1^(2) = 1 and RHS = 1/6 xx 1 xx 2(2xx 1+1) = 1`. ` :." " `LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1^(2)+2^(2)+3^(2)+...+k^(2)= 1/6 k(k+1)(2k+1)`. ...(i) Now, `1^(2)+2^(2)+3^(2)+...+k^(2)+(k+1)^(2)` ` = {1^(2)+2^(2)+3^(2)+...+k^(2)}+(k+1)^(2)` ` = 1/6 k(k+1)(2k+1)+(k=1)^(2)" "` [using (i)] `=1/6 (k+1)*{k(2k+1)+6(k+1)}` ` = 1/6 (k+1)(2k^(2)+7k+6) = 1/6 (k+1){(2k^(2)+4k)+(3k+6)}` ` = 1/6 (k+1)(k+2)(2k+3)= 1/6 (k+1)(k+1)[2(k+1)+1]`. ` :." " P(k+1):1^(2)+2^(2)+3^(2)+...+(k+1)^(2)` ` " "= 1/6 (k+1)(k+1+1)[2(k+1)+1]`. This shows that P(k+1) is true, whenever P(k) is true. Thus, P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, we have ` 1^(2)+2^(2)+3^(2)+...+n^(2) = 1/6 n(n+1) (2n+1)" for all " n in N`.

Prove by the principle of mathematical induction that for all `n in N :``1^2+2^2+3^2++n^2=1/6n(n+1)(2n+1)`