JEE Mathematics 2025 Question with Solution

Given: The series is $1 + 3 + 5^2 + 7 + 9^2 + 11 + 13^2 + \cdots$ and we need the sum of the first $40$ terms.

Find: The total sum of these $40$ terms.

Terms alternate between plain odd numbers and squared odd numbers, with the first two terms $1$ and $3$ plain. Therefore, in the first $40$ terms:

• plain odd terms = $21$ terms
• squared odd terms = $19$ terms

The plain terms are $1$ and $3, 7, 11, \ldots, 79$. So,

$$S_{\text{plain}} = 1 + \frac{20}{2}(3+79)$$ $$= 1 + 10 \cdot 82 = 821$$

The squared terms are $5^2, 9^2, 13^2, \ldots, 77^2$. Their bases are $4k+1$ for $k=1,2,\ldots,19$. Thus,

$$S_{\text{sq}}=\sum_{k=1}^{19}(4k+1)^2$$ $$= \sum_{k=1}^{19}(16k^2+8k+1)$$ $$= 16\sum_{k=1}^{19}k^2 + 8\sum_{k=1}^{19}k + 19$$

Using

$$\sum_{k=1}^{19}k = \frac{19\cdot 20}{2} = 190$$

and

$$\sum_{k=1}^{19}k^2 = \frac{19\cdot 20\cdot 39}{6} = 2470$$

we get

$$S_{\text{sq}} = 16\cdot 2470 + 8\cdot 190 + 19$$ $$= 39520 + 1520 + 19 = 41059$$

Therefore,

$$S_{40} = S_{\text{plain}} + S_{\text{sq}} = 821 + 41059 = 41880$$

So, the correct option is B.

Given: The series is $1 + 3 + 5^2 + 7 + 9^2 + \ldots$ upto $40$ terms.

Find: Its total sum.

Separate the series into two parts:

1. terms of the form odd number
2. terms of the form square of an odd number

From the pattern used in the solution, the first $40$ terms contain $21$ plain odd terms and $19$ squared odd terms.

For plain terms:

$$1, 3, 7, 11, \ldots, 79$$

So,

$$S_{\text{plain}} = 1 + \frac{20}{2}(3+79)$$ $$= 1 + 820 = 821$$

For squared terms:

$$5^2, 9^2, 13^2, \ldots, 77^2$$

Write these as

$$(4k+1)^2, \quad k=1 \text{ to } 19$$

Hence,

$$\sum_{k=1}^{19}(4k+1)^2 = \sum_{k=1}^{19}(16k^2+8k+1)$$ $$= 16\sum_{k=1}^{19}k^2 + 8\sum_{k=1}^{19}k + \sum_{k=1}^{19}1$$ $$= 16\cdot 2470 + 8\cdot 190 + 19$$ $$= 39520 + 1520 + 19 = 41059$$

Now add both parts:

$$821 + 41059 = 41880$$

Therefore, the sum upto $40$ terms is $41880$, so the correct option is B.