Animated Solution for Mathematics - Binomial Theorem: Let $S_1={\sum }_{j=1}^{10}j(j-1{)}^{10}{C}_j,S_2={\sum }_{j=1}^{10}{j}^{10}{C}_j$ and S_3 = \sum_{j=1}^{10} j^2 ^{10}C_j. Statement-1: $S_3=55\times 2^9$. Statement-2: $S_1=90\times 2^8$ and $S_2=10\times 2^8$.

Imagine you are standing on the precipice of a complex summation. You see terms like $j(j-1)\left(\genfrac{}{}{0ex}{}{10}{j}\right)$ and $j^2\left(\genfrac{}{}{0ex}{}{10}{j}\right)$. At first glance, they look intimidating, perhaps even chaotic.

But in the world of JEE Advanced, chaos is just order waiting to be discovered. Today, we are going to peel back the layers of these sums and reveal the beautiful, underlying structure of binomial coefficients.

Let us begin with $S_2={\sum }_{j=1}^{10}j\left(\genfrac{}{}{0ex}{}{10}{j}\right)$. This is the heartbeat of our problem.

We rely on the standard identity:

When we substitute $n=10$, we get $S_2=10\times 2^{10-1}=10\times 2^9$. Notice how the power of $2$ is $n-1$. If you ever find yourself writing $2^8$ here, pause! That is the trap.

The identity is robust, and our calculation gives us $10\times 2^9$. Statement-2 claims $S_2=10\times 2^8$, which we now know is false. The first domino has fallen.

Now, consider $S_1={\sum }_{j=1}^{10}j(j-1)\left(\genfrac{}{}{0ex}{}{10}{j}\right)$. This looks more complex, but it is actually a gift.

We have a specific identity for this:

Substituting $n=10$, we calculate $S_1=10(9)2^{10-2}=90\times 2^8$. This matches Statement-2 perfectly.

However, because the $S_2$ part of Statement-2 was incorrect, the entire statement is false. Do not let the partial truth distract you from the logical conclusion!

Decomposing $j^2$

Finally, we arrive at $S_3={\sum }_{j=1}^{10}{j}^2\left(\genfrac{}{}{0ex}{}{10}{j}\right)$. We have no direct formula for $j^2\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, but we have a secret weapon: algebraic decomposition.

We can rewrite $j^2$ as $j(j-1)+j$. Why? Because $j(j-1)$ perfectly matches the structure of $S_1$, and $j$ matches $S_2$.

Thus, we have:

Now, we bring it all together. We have $S_1=90\times 2^8$ and $S_2=10\times 2^9$. To add these, we need a common power of $2$.

Let us express $10\times 2^9$ as $20\times 2^8$. Now the addition is trivial:

Factoring out a $2$, we get $55\times 2\times 2^8=55\times 2^9$. This matches Statement-1 exactly.

We have navigated the complexity, avoided the traps, and arrived at the truth: Statement-1 is True, and Statement-2 is False. Remember, in JEE, the most complex problems are often just simple identities wearing a disguise. Keep practicing, and you will see through the disguise every time.