Question

1 + 3 + $5^2$ + 7 + $9^2$ + $\ldots$ upto 40 terms is equal to

• A. 43890
• B. 41880
• C. 33980
• D. 40870

Hint:

Separate the series into parts and sum each part individually.

Answer 1

The Correct Option is B

1. Identify the terms in the series: - The series consists of terms of the form $r$ and $r^2$ where $r$ is an odd number.
2. Separate the series into two parts: - Part 1: Sum of terms of the form $r$. - Part 2: Sum of terms of the form $r^2$.
3. Sum of terms of the form $r$: - The sequence is $1, 3, 5, 7, \ldots$ up to 20 terms. - Sum of the first 20 odd numbers: \[ \sum_{r=1}^{20} (2r-1) = 20^2 = 400 \]
4. Sum of terms of the form $r^2$: - The sequence is $1^2, 3^2, 5^2, 7^2, \ldots$ up to 20 terms. - Sum of the squares of the first 20 odd numbers: \[ \sum_{r=1}^{20} (2r-1)^2 = \sum_{r=1}^{20} (4r^2 - 4r + 1) \] \[ = 4 \sum_{r=1}^{20} r^2 - 4 \sum_{r=1}^{20} r + \sum_{r=1}^{20} 1 \] \[ = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 4 \cdot \frac{20 \cdot 21}{2} + 20 \] \[ = 4 \cdot 2870 - 4 \cdot 210 + 20 = 11480 - 840 + 20 = 10660 \] 5. Total sum of the series: \[ \text{Total sum} = 400 + 10660 = 41880 \] Therefore, the correct answer is (2) 41880.

Answer 2

We need the sum of the first 40 terms of the series: \(1 + 3 + 5^2 + 7 + 9^2 + 11 + 13^2 + \cdots\).

Concept Used:

Terms alternate between a plain odd number and a squared odd number, except the very first two terms (1 and 3) are both plain. Thus, among the first 40 terms:

• Plain (unsquared) terms: 21 terms (the first term 1, plus the 20 even-position terms).
• Squared terms: 19 terms (odd positions from 3 to 39).

Step-by-Step Solution:

Step 1: Sum of the 21 plain odd terms.

These are: \(1\) and \(3,7,11,\ldots,79\) (20 terms in an AP with first term 3 and common difference 4).

\[ S_{\text{plain}} = 1 + \frac{20}{2}\,(3+79) = 1 + 10 \cdot 82 = 1 + 820 = 821. \]

Step 2: Sum of the 19 squared terms.

The squared bases are \(5,9,13,\ldots,77\), i.e., \(4k+1\) for \(k=1,\ldots,19\). Hence

\[ S_{\text{sq}}=\sum_{k=1}^{19}(4k+1)^2 =\sum_{k=1}^{19}\big(16k^2+8k+1\big) =16\sum_{k=1}^{19}k^2 + 8\sum_{k=1}^{19}k + 19. \] \[ \sum_{k=1}^{19}k = \frac{19\cdot 20}{2} = 190,\qquad \sum_{k=1}^{19}k^2 = \frac{19\cdot 20\cdot 39}{6} = 2470. \] \[ S_{\text{sq}} = 16\cdot 2470 + 8\cdot 190 + 19 = 39520 + 1520 + 19 = 41059. \]

Step 3: Total sum.

\[ S_{40} = S_{\text{plain}} + S_{\text{sq}} = 821 + 41059 = \boxed{41880}. \]

Final Computation & Result

Answer: 41880