Question

The sum of the first n terms of the series \[ 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \cdots \] {is} \[ \frac{n(n + 1)^2}{2} { when n is even. When n is odd the sum is} \]

• A. \( \left[ \frac{n(n + 1)}{2} \right]^2 \)
• B. \( \frac{n^2(n + 1)}{2} \)
• C. \( \frac{n(n + 1)^2}{4} \)
• D. \( \frac{3n(n + 1)}{2} \)

Hint:

The sum of squares of consecutive terms with a pattern, like this one, can often be separated into two different series to make the calculation easier.

Answer 1

The Correct Option is B

Step 1: The given series is of the form \[ S = 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \cdots \] For even terms, the general pattern is \( n^2 \), and for odd terms, the general pattern is \( 2n^2 \). 
Step 2: The sum of the first n terms can be separated into two series. One series corresponds to even terms and the other to odd terms. 
Step 3: By using the summation formula for squares, the sum for even \( n \) results in the expression \[ S = \frac{n^2(n+1)}{2} \]