Prove that `1^2+2^2+dotdotdot+n^2>(n^3)/3,``n in N`

1.	Prove that `1^2+2^2+dotdotdot+n^2>(n^3)/3,``n in N`
Answer» We will prove it using mathematical induction. For, `n = 1`, `L.H.S = 1^2 = 1` `R.H.S. = 1^3/3 = 1/3` As, `1 gt 1/3`, our equation is true for `n = 1`. Let, our equation is true for `n = k` where `k` is a natural number. Then, `1^2+2^2+3^2+...k^2 gt k^3/3->(1)` Now, we have to prove, for `n = k+1`, given equation is true. For, `n = k+1`, `L.H.S. = 1^2+2^2+3^2+...k^2+(k+1)^2` From (1), `1^2+2^2+3^2+...k^2+(k+1)^2 gt k^3/3+(k+1)^2` `gt 1/3(k^3+3k^2+6k+3)` `gt 1/3((k^3+1^3+3k(k+1))+(3k+2))` `gt 1/3((k+1)^3)+(3k+2))` `gt 1/3(k+1)^3+1/3(3k+2)` `gt 1/3(k+1)^3`, as `1/3(3k+2) gt 0` `:. 1^2+2^2+3^2+...k^2+(k+1)^2 gt 1/3((k+1)^3)` Thus, our equation is true for `n = k+1`. `:. 1^2+2^2+3^2+...n^2 gt n^3/3`

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