(1^2 + 2^2 + …… + n^2) ........... for all values of n N:

MCQ

$(1^2 + 2^2 + …… + n^2) ...........$ for all values of $n \in N:$

A
$= n^3/3$
B
$< n^3 /3$
✓
$> n^3/3$
D
None of these

✓

Answer

Correct option: C.

$> n^3/3$

Let $P(n): (1^2 + 2^2 + ….. + n^2) > n^3/3.$
When $= 1, \text{LHS} = 1^2 = 1$ and $\text{RHS} = 1^3/3 = 1/3.$
Since $1 > 1/3,$ it follows that $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): (1^2 + 2^2+ ….. + k^2 ) > k^3/3 …. (i)$
Now,
$1^2 + 2^2 + ….. + k^2+ (k + 1)^2$
$= \{1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> k^3/3 + (k + 1)^3 [$using $(i)]$
$= 1/3 ∙ (k^3 + 3 + (k + 1)^2) = 1/3 ∙ \{k^2 + 3k^2 + 6k + 3\}$
$= 1/3[k^3 + 1 + 3k(k + 1) + (3k + 2)]$
$= 1/3 ∙ [(k + 1)^3 + (3k + 2)]$
$> 1/3(k + 1)^3P(k + 1):$
$1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> 1/3 ∙ (k + 1)^3$
$P(k + 1)$ is true, whenever $P(k)$ is true.
Thus $P(1)$ is true and $P(k + 1)$ is true whenever $p(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$

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