JEE Main 2025 — Sequences And Series Question with Solution

Step 1: Breaking the sequence into two parts

The given sequence is $1+3+5^2+7+9^2+\dots$ up to 40 terms. Let's group the terms as follows:

The terms are arranged like this: $1$, $3$, $5^2$, $7$, $9^2$, $11$, $13^2$, $15$, $17^2$, ... You can see that every third term is being squared, while others are just numbers. So, split the sequence into: all squared terms ($1^2,5^2,9^2,...$) and all other numbers ($3,7,11,...$).

Step 2: Counting the terms in each group

Out of 40 terms, half will be squared terms and half will be non-squared terms. So we have 20 squared terms and 20 ordinary numbers.

Step 3: Sums for each group

The squared numbers follow this pattern: $1^2,5^2,9^2,...$, which can be written as $(4k-3{)}^2$ for $k$ from 1 to 20.

The other numbers are $3,7,11,...,(4k-1)$ for $k$ from 1 to 20.

Step 4: Write these as formulas

Total sum = Sum of squares part $+$ Sum of ordinary numbers part.

The sum of the squared numbers is ${\sum }_{k=1}^{20}(4k-3{)}^2$.

The sum of the other numbers (arithmetic series): $3+7+11+...$ for 20 terms.

The first term is 3, the last term is $3+(20-1)\times 4=3+76=79$. So, the sum is $\frac{20}{2}[3+79]=10\times 82=820$.

Step 5: Expanding and simplifying

Expand $(4k-3{)}^2$: $$(4k-3{)}^2=16k^2-24k+9$$

So, sum up these for $k=1$ to $20$: $16\sum k^2-24\sum k+9\times 20$

We know formulas: ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$
${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$

Plug in $n=20$: ${\sum }_{k=1}^{20}{k}^2=\frac{20\times 21\times 41}{6}$
${\sum }_{k=1}^{20}k=\frac{20\times 21}{2}$

Now, sum up:

$=16\left(\frac{20\times 21\times 41}{6}\right)-24\left(\frac{20\times 21}{2}\right)+180+820$

Step 6: Calculating

$16\left(\frac{20\times 21\times 41}{6}\right)=16\times 2870=45920$
$24\left(\frac{20\times 21}{2}\right)=24\times 210=5040$
$9\times 20=180$
Sum of the other numbers = $820$

So, total = $45920-5040+180+820$
$=45920-5040=40880$
$40880+180=41060$
$41060+820=41880$

Final Answer

The sum of the 40 terms is $41880$.