Animated Solution for Mathematics - Sequence and Series: The sum $1^2-2.3^2+3.5^2-4.7^2+5.9^2-\cdots +15.29^2$ is

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a series that, at first glance, looks like a chaotic mess of alternating signs and squared terms.

The problem asks us to compute the sum $S=1^2-2\cdot 3^2+3\cdot 5^2-4\cdot 7^2+5\cdot 9^2-\cdots +15\cdot 29^2$.

When you see a series like this, your first instinct might be to reach for a calculator or start grinding out the arithmetic. Stop. Take a breath. In JEE Advanced, the secret is never brute force; it is always structure.

Decoding the DNA of the Series

Look closely at the terms. The coefficients are $1,2,3,4,\dots ,15$. The bases of the squared terms are $1,3,5,7,\dots ,29$, which are consecutive odd numbers.

If we denote the index as $n$, the $n^{th}$ coefficient is simply $n$. The $n^{th}$ odd number is $2n-1$. The alternating sign is handled by $(-1{)}^{n+1}$.

Thus, our general term is $T_n=(-1{)}^{n+1}\cdot n(2n-1{)}^2$. The entire sum is:

This is the DNA of our problem.

The Strategy of Grouping

Dealing with alternating signs in a summation is like trying to walk on a floor that keeps changing its elevation. Let us define a function $f(n)=n(2n-1{)}^2$.

Now, our sum is $S=f(1)-f(2)+f(3)-f(4)+\cdots +f(15)$. Instead of fighting the signs, let us embrace them by grouping.

We keep $f(1)$ separate and pair the rest:

This condenses into $S=f(1)+{\sum }_{k=1}^7[f(2k+1)-f(2k)]$. This is the moment where the complexity vanishes.

The Algebraic Engine

Now, we need to expand $f(n)=n(2n-1{)}^2$. Expanding the square gives $f(n)=n(4n^2-4n+1)$, which simplifies to $f(n)=4n^3-4n^2+n$.

To find the difference $f(n+1)-f(n)$, we calculate:

Using the expansion formulas, this simplifies to $\Delta f=12n^2+4n+1$. We need this for $n=2k$, so substituting $n=2k$ gives:

The Final Calculation

We are almost there. Our sum is $S=f(1)+{\sum }_{k=1}^7(48k^2+8k+1)$. We know $f(1)=1(2(1)-1{)}^2=1$.

The summation splits into three parts:

Using the standard formulas ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$ and ${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$, we calculate for $n=7$:

Finally, substituting these values:

The elegance of this cancellation is the true reward of the JEE process. Keep this technique in your toolkit—it is a lifesaver for alternating series! The final answer is 6952.