Question

By PMI , Prove that
1^2 + 2^2 + ... + n^2 >n^3÷3 n€N ​

Answer 1

$</p><p>\huge{ \underline{ \boxed{ \mathfrak{ \red{question}}}}} \\ \\ {1}^{2} + {2}^{2} + {3}^{2} + ......... {n}^{2} > \frac{ {n}^{3} }{3}$

$</p><p>\huge{ \underline{ \boxed{ \mathfrak{ \red{answer}}}}}$

We have to prove that

1^2 + 2^2 + ... + n^2 >n^3÷3 n€N........(1)

put n=1

${1}^{2} > ( \frac{1}{3})^{3} \\ \\ 1 > \frac{1}{3}$

which is true.

∴Result is true for n=k

$∴ {1}^{2} + {2}^{2} + {3}^{2} + .... + {k}^{2} > \frac{ {k}^{3} }{3} \\ \\∴ {1}^{2} + {2}^{2} + {3}^{2} + .... + {k}^{2} + ( {k + 1})^{2} > \frac{ {k}^{3} }{3} + ( {k + 1) }^{2} \\ \\ = > \frac{ {k}^{3} + 3( {k + 1)}^{2} }{3} = \frac{ {k}^{3} + 3 {k}^{2} + 6k + 3 }{3} \\ \\ = > \frac{ {k}^{3} + 3 {k}^{2} + 3k + 1 + 3(k + 2)}{3} = \frac{( {n + 1)}^{2} }{3} + (k + \frac{2}{3} ) \\ \\ ∴ {1}^{2} + {2}^{2} + {3}^{2} + ...... + {k}^{2} + ( {k + 1)}^{2} > \frac{ ({k + 1)}^{3} }{3} </p><p>$

Result is true for n= k+1

Hence by mathematical induction P(n)

is true for all n€N.