Explanation
1. Define Ak:
The arithmetic mean of the squares of the first k natural numbers is given by:
Ak=k12+22+⋯+k2
Using the formula for the sum of the first k squares, ∑i=1ki2=6k(k+1)(2k+1), we get:
Ak=k6k(k+1)(2k+1)=6(k+1)(2k+1)
2. Simplify the term (6Ak−3k):
Substitute the expression for Ak:
6Ak−3k=6[6(k+1)(2k+1)]−3k
=(k+1)(2k+1)−3k
=(2k2+3k+1)−3k
=2k2+1
3. Evaluate the summation:
We are given ∑k=1n(6Ak−3k)=31. Substituting our simplified expression:
k=1∑n(2k2+1)=31
2k=1∑nk2+k=1∑n1=31
2[6n(n+1)(2n+1)]+n=31
3n(n+1)(2n+1)+n=31
4. Solve for n:
Testing integer values for n:
Since the equation holds true for n=3, the value is 3.
Correct Option: 3. 3