Question

prove that 1^2+2^2+...+n^2 is greater than n^3/3,n belongs to N.??​

Answer 1

Answer:

Let P(n) be the statement given by

P(n) = 1^2 + 2^2 + 3^2.... +n^2 > n^3/3

P(1 ) = L.H.S = 1^2 > 1^3/3

=> 1^2 = 1>1/3 = 1^3/3

P(1) is true.

Now let us show that P(k) is true for some positive integer integer k.

P(k) = 1^2 + 2^2 + 3^2 +.... + k^2 > k^3/3

We shall now prove that P( k+1) is also true.

P( k + 1 ) = 1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > ( k+1) ^3/3

Since, P(k) is true.

Adding ( m + 1) ^2 in both sides of P( k ),

1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > (k^3/ 3) + ( k +1 ) ^2

1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > ( k^3/3) + ( k^2 + 1 + 2k )

1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > 1/3 ( k^3 + 3k^2 + 3 + 6k)

1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > 1/3 {( k^3 + 3k^2 + 3k) + ( 3k + 2)}

1^2 + 2^2 + 3^2 +.... + k^2 + ( k + 1) ^2 > 1/3 { ( k + 1) ^3 + ( 3m +2)} > ( m + 1) ^3 / 3

=> P( k +1) is also true.

By PMI, P(n) is true for all n belongs to N.