Question

Let \(A_k\) be the arithmetic mean of the squares of \(k\) natural numbers. \[ \sum_{k=1}^{n}(6A_k-3k)=31 \] Find the value of \(n\):

• A. \(3\)
• B. \(2\)
• C. \(4\)
• D. \(1\)

Hint:

Whenever the arithmetic mean of squares is involved, immediately use \[ 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}. \] This formula simplifies most summation problems directly.

Answer 1

The Correct Option is A

Concept: The arithmetic mean of the squares of the first \(k\) natural numbers is \[ A_k=\frac{1^2+2^2+\cdots+k^2}{k} \] Using the standard formula \[ 1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6} \] we can simplify \(A_k\) and then evaluate the given summation.

Step 1: Find the expression for \(A_k\).
\[ A_k=\frac{1}{k}\cdot\frac{k(k+1)(2k+1)}{6} \] \[ A_k=\frac{(k+1)(2k+1)}{6} \] Therefore, \[ 6A_k=(k+1)(2k+1) \] \[ =2k^2+3k+1 \] Hence \[ 6A_k-3k=2k^2+1 \]

Step 2: Substitute into the given summation.
\[ \sum_{k=1}^{n}(6A_k-3k) = \sum_{k=1}^{n}(2k^2+1) \] \[ = 2\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}1 \] \[ = 2\cdot\frac{n(n+1)(2n+1)}{6}+n \] \[ = \frac{n(n+1)(2n+1)}{3}+n \] Given that this equals \(31\), \[ \frac{n(n+1)(2n+1)}{3}+n=31 \]

Step 3: Solve for \(n\).
Multiplying by \(3\), \[ n(n+1)(2n+1)+3n=93 \] Checking \(n=2\), \[ 2(3)(5)+6=30+6=36 \] \[ \frac{36}{3}=12 \] Checking \(n=3\), \[ 3(4)(7)+9=84+9=93 \] \[ \frac{93}{3}=31 \] Thus, \[ n=3 \] \[ \boxed{n=3} \]