Question

For \(n\in\mathbb{N}\), \[ 1^2+2^2+3^2+\cdots+n^2 > \]

• A. \(n^3\)
• B. \(\dfrac{n^3}{2}\)
• C. \(\dfrac{n^3}{3}\)
• D. \(3n^3\)

Hint:

Remember the important identity: \[ 1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6} \] It is frequently used in inequalities and estimations.

Answer 1

The Correct Option is C

Concept: The sum of squares of first \(n\) natural numbers is: \[ 1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6} \] We compare this expression with the given options.

Step 1: Write the standard formula.
\[ S= \frac{n(n+1)(2n+1)}{6} \] Expand numerator: \[ (n+1)(2n+1)=2n^2+3n+1 \] Thus: \[ S= \frac{2n^3+3n^2+n}{6} \]

Step 2: Compare with \(\dfrac{n^3}{3}\).
Observe that: \[ S= \frac{2n^3}{6}+\frac{3n^2+n}{6} \] \[ S= \frac{n^3}{3}+\frac{3n^2+n}{6} \] Since: \[ \frac{3n^2+n}{6}>0 \] for all \(n\in\mathbb{N}\), we get: \[ S>\frac{n^3}{3} \]

Step 3: Check other options.
The inequality: \[ S>\frac{n^3}{2} \] is not always true for all \(n\). Similarly: \[ S>n^3 \] and \[ S>3n^3 \] are false. Thus only option (C) is always true.

Step 4: Choose the correct answer.
Hence: \[ \boxed{ 1^2+2^2+\cdots+n^2>\frac{n^3}{3} } \]