Animated Solution for Mathematics - Sequence and Series: Let $S_n={\sum }_{k=1}^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2$. Then $S_n$ can take value(s)

Welcome, future engineer! Today, we are going to look at a problem that, at first glance, seems like a chaotic mess of alternating signs and squares. You see the expression $S_n={\sum }_{k=1}^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2$ and your instinct might be to panic.

But take a deep breath. In mathematics, especially in the JEE, complexity is often just a mask for hidden symmetry. Our job is to pull off that mask.

Let us look at the exponent $\frac{k(k+1)}{2}$. This is the formula for the $k$-th triangular number. Does it behave randomly?

Let us test the first few values of $k$: For $k=1$, the exponent is $1$, so the sign is $(-1{)}^1=-1$. For $k=2$, the exponent is $3$, so the sign is $(-1{)}^3=-1$. For $k=3$, the exponent is $6$, so the sign is $(-1{)}^6=+1$. For $k=4$, the exponent is $10$, so the sign is $(-1{)}^{10}=+1$.

Look at that! The pattern is $-,-,+,+$. If you continue for $k=5,6,7,8$, you will find the pattern repeats exactly. We have discovered a cycle of length $4$.

Since the sum goes up to $4n$, we have exactly $n$ blocks of $4$ terms. This is the breakthrough we needed.

Now, let us define the $r$-th block, which we will call $T_r$. This block contains the terms from $k=4r-3$ to $k=4r$. Writing it out, we get:

Instead of expanding these squares—which would be a nightmare of algebra—let us use the difference of squares identity: $a^2-b^2=(a-b)(a+b)$. We can rearrange $T_r$ as:

Applying the identity to the first pair:

Applying it to the second pair:

Adding these together, the expression for our block simplifies beautifully to $T_r=32r-12$. We have turned a complex series of squares into a simple linear expression.

Now, the total sum $S_n$ is just the sum of these $n$ blocks:

We can split this into two parts:

Using the standard formula ${\sum }_{r=1}^{n}r=\frac{n(n+1)}{2}$, we get:

Expanding this, we arrive at our final, powerful result: $S_n=16n^2+4n$.

We have the formula. Now we just need to test it against the options. If we plug in $n=8$, we get $S_8=16(64)+32=1024+32=1056$.

If we plug in $n=9$, we get $S_9=16(81)+36=1296+36=1332$. Both of these values appear in our options!

Remember, the goal isn't just to get the answer; it is to understand the journey. You started with a daunting summation and, through pattern recognition and algebraic manipulation, you tamed it. Keep this mindset, and there is no problem in the JEE that you cannot solve.