Using the principle of mathematical induction. Prove that `(1^(2)+2^(2)+…+n^(2)) gt n^(3)/3 " for all values of " n in N`.

Using the principle of mathematical induction. Prove that `(1^(2)+2^(2

1.	Using the principle of mathematical induction. Prove that `(1^(2)+2^(2)+…+n^(2)) gt n^(3)/3 " for all values of " n in N`.
Answer» Let `P(n): (1^(2)+2^(2)+…+n^(2)) gt n^(3)/3`. When n =1, LHS = `1^(2) = 1 and RHS = 1^(3)/3 = 1/3`. Since ` 1 gt 1/3`, it follows that P(1) is true. Let P(k) be true. Then, `P(k): (1^(2)+2^(2)+…+k^(2)) gt k^(3)/3`. ….(i) Now, ` 1^(2)+2^(2)+...+k^(2)+(k+1)^(2) ` ` ={1^(2)+2^(2)+...+k^(2)}+(k+1)^(2)` ` gt k^(3)/3 + (k+1)^(2)` [using (i)] ` = 1/3{k^(3)+3(k+1)^(2)} = 1/3 {k^(3)+3k^(2)+6k+3}` ` = 1/3 [{(k^(3)+1+3k(k+1)}+(3k+2)]=1/3*[(k+1)^(3) +(3k+2)]` ` gt 1/3 (k+1)^(3)`. ` :. P(k+1): 1^(2)+2^(2)+...+k^(2)+(k+1)^(2) gt 1/3 (k+1)^(3)`. Thus, P(k+1) is true, whenever P(k) is true. `:. ` P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, it follows that `(1^(2)+2^(2)+3^(2)+...+n^(2)) gt n^(3)/3 " for all " n in N`.