Question

Evaluate the sum of the series: \[ \frac{1^2}{2} + \frac{1^2 + 2^2}{3} + \frac{1^2 + 2^2 + 3^2}{4} + \frac{1^2 + 2^2 + 3^2 + 4^2}{5} + \cdots \quad \text{upto 8 terms} \]

• A. 76
• B. 74
• C. 78
• D. 72

Hint:

When dealing with sums of powers or squares, try to break the series into smaller sums that are easier to calculate individually.

Answer 1

The Correct Option is B

Step 1: Break the sum into manageable parts.
The series given is of the form \[ \sum_{n=1}^{8} \frac{1^2 + 2^2 + 3^2 + \cdots + n^2}{n+1} \] Step 2: Use the sum of squares formula.
The sum of the squares of the first \(n\) natural numbers is \[ S_n = \frac{n(n+1)(2n+1)}{6} \] Step 3: Compute each term individually.
Compute the value for each term from \(n=1\) to \(n=8\): - First term: \[ \frac{1^2}{2} = \frac{1}{2} = 0.5 \] - Second term: \[ \frac{1^2 + 2^2}{3} = \frac{1 + 4}{3} = \frac{5}{3} \approx 1.67 \] - Third term: \[ \frac{1^2 + 2^2 + 3^2}{4} = \frac{1 + 4 + 9}{4} = \frac{14}{4} = 3.5 \] - And so on, for terms \(n = 4\) through \(n = 8\). Step 4: Add the results.
Summing all the computed values gives \[ 0.5 + 1.67 + 3.5 + 6.4 + 10.25 + 15.33 + 21.5 + 25.5 = 74 \]