To find: 12+22+32...................+n2

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Question

To find:

1^{2} + 2^{2} + 3^{2} . . . . . . . . . . . . . . . . . . . + n^{2}

Options:

A . =n32

B . >n32

C . >n33

D . =n33

Answer: Option C
:
C
Let

n = 1

\frac{n^{3}}{2} = \frac{1}{2}; \frac{n^{3}}{3} = \frac{1}{3}

Let

n = 2, \Rightarrow 1^{2} + 2^{2} = 5

\frac{n^{3}}{2} = 4; \frac{n^{3}}{3} = \frac{8}{3}

Let

n = 3, \Rightarrow 1^{2} + 2^{2} + 3^{2} = 14

\frac{n^{3}}{2} = \frac{27}{2}; \frac{n^{3}}{3} = 9

Let

n = 4, \Rightarrow 1^{2} + 2^{2} + 3^{2} + 4^{2} = 30

\frac{n^{3}}{2} = 32; \frac{n^{3}}{3} = \frac{64}{3}

1^{2} + 2^{2} + 3^{2} + \dots + n^{2} > \frac{n^{3}}{3}

P (n) : 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} > \frac{n^{3}}{3}

P (1)

is true
Let

P (k)

be true.

\Rightarrow 1^{2} + 2^{2} + 3^{2} + \dots + k^{2} > \frac{k^{3}}{3}

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} > \frac{k^{3}}{3} + k^{2} + 2 k + 1

= \frac{k^{3} + 3 k^{2} + 6 k + 3}{3}

= \frac{k^{3} + 3 k^{2} + 3 k + 1 + 3 k + 2}{3}

= \frac{(k + 1)^{3}}{3} + k + \frac{2}{3}

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} > \frac{(k + 1)^{3}}{3}

P (k + 1)

is true.

P (n)

is true

\forall n \in N

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