1. Identify the pattern of terms

The series is $1+3+5^2+7+9^2+\ldots$

This means:

• odd numbers appear in order: $1,3,5,7,9,\ldots$
• every alternate odd term is squared.

So the terms are: $1,\;3,\;5^2,\;7,\;9^2,\;11,\;13^2,\ldots$

Thus:

• odd-positioned terms are unsquared odd numbers,
• even-positioned terms are squares of odd numbers starting from $5$?
  Actually from the given pattern, the sequence is better grouped as: $1,3,5^2,7,9^2,11,13^2,15,\ldots$ so every third? No. Let us rewrite carefully.

Observe the odd numbers in order: $1,3,5,7,9,11,13,15,\ldots$ Among these, numbers congruent to $1 \pmod 4$ after the first two are not squared, while numbers congruent to $1 \pmod 4$? This is messy.

A more natural interpretation used in such problems is: $1+3+5^2+7+9^2+\cdots$ means the odd numbers are taken in order and every term from $5$ onward alternates between squared and unsquared: $1,3,5^2,7,9^2,11,13^2,15,\ldots$

So among 40 terms:

• 20 terms are unsquared odd numbers: $1,3,7,11,15,\ldots$
• 20 terms are squared odd numbers: $5^2,9^2,13^2,\ldots$

But let us index directly.

2. Write the two subsequences

The 40 terms split into:

• 20 unsquared terms: $1,3,7,11,15,\ldots,79$ This is not arithmetic from the beginning because of the first jump, so this interpretation becomes inconsistent.

Hence the intended meaning must be different.

---

3. Correct interpretation

The expression is most reasonably read as: $1+3^2+5+7^2+9+\cdots$ would alternate square status, but that is not what is printed.

So let us inspect the options using the standard pattern often intended in such JEE-style questions:

The odd numbers are: $1,3,5,7,9,\ldots,79$ for 40 terms.

Among them, the terms in positions $3,5,7,\ldots,39$ are squared, i.e. 19 squared terms, and the remaining 21 are unsquared.

Then:

• squared numbers are $5,9,13,\ldots,77$ i.e. an AP with 19 terms, general term $4k+1$ for $k=1$ to $19$.
• unsquared numbers are $1,3,7,11,15,\ldots,79$.

Compute both sums.

4. Sum of squared terms

Squared terms: $5^2+9^2+13^2+\cdots+77^2$ General term: $4k+1, \quad k=1 \text{ to } 19$ So $S_1=\sum_{k=1}^{19}(4k+1)^2$ $=(16\sum k^2 + 8\sum k + \sum 1)_{1}^{19}$

Now, $\sum_{k=1}^{19} k = \frac{19\cdot20}{2}=190$ $\sum_{k=1}^{19} k^2 = \frac{19\cdot20\cdot39}{6}=2470$

Thus, $S_1=16(2470)+8(190)+19$ $=39520+1520+19=41059$

5. Sum of unsquared terms

All 40 odd numbers sum to $1+3+5+\cdots+79=40^2=1600$

Among these, the numbers that were squared instead of taken normally are $5+9+13+\cdots+77$ So unsquared contribution is $S_2=1600-(5+9+13+\cdots+77)$

Now, $5+9+13+\cdots+77$ has 19 terms, AP with first term $5$, last term $77$.

So, $\text{sum}=\frac{19}{2}(5+77)=\frac{19}{2}\cdot82=19\cdot41=779$

Therefore, $S_2=1600-779=821$

6. Total sum

$S=S_1+S_2=41059+821=41880$

7. Compare with options

$41880$ which matches Option B.

8. Comparison with stored answer

Stored correct answer is B, which agrees with the derived result.