Animated Solution for Mathematics - Sequence and Series: Let $S_k=\frac{1+2+3+...+k}{k}$. If $S_1^2+S_2^2+...+S_{10}^2=\frac{5}{12}A$, then A is equal to :

Unlocking the Series

Have you ever looked at a complex-looking series problem and felt that immediate, sinking sensation of dread? You see a summation, a fraction, and a series inside a series, and your brain starts screaming, "This is going to take forever!"

But here is the secret that separates the top rankers from the rest: the most intimidating problems are often the ones that collapse the fastest if you just know where to apply the pressure.

Today, we are going to dismantle a classic JEE-style series problem. We are given $S_k=\frac{1+2+3+...+k}{k}$, and we need to find the value of $A$ in the equation ${\sum }_{k=1}^{10}{S}_k^2=\frac{5}{12}A$. Let's take a breath and dive in.

The Power of the General Term

The biggest mistake students make is jumping straight into the summation. They see $S_1^2+S_2^2+...+S_{10}^2$ and start trying to calculate each term individually. Please, never do that!

The key to any series problem is to simplify the general term, $S_k$, before you even think about the summation symbol.

Look at the numerator: $1+2+3+...+k$. This is the sum of the first $k$ natural numbers. We know this formula by heart:

Now, let's substitute this back into our expression for $S_k$. Instead of that bulky numerator, we now have:

Look at that! The $k$ in the numerator and the $k$ in the denominator are just waiting to be canceled. Since $k$ represents the position in the sequence, it starts at $1$ and goes to $10$, so $k$ is never zero.

We can safely cancel it out. Suddenly, our terrifying expression $S_k$ simplifies to something beautiful and elegant:

The Elegance of Index Shifting

Now that we have $S_k=\frac{k+1}{2}$, the problem asks for the sum of the squares, $S_k^2$. Squaring our simplified term is trivial:

We need to evaluate ${\sum }_{k=1}^{10}{S}_k^2$. Let's pull that constant $\frac{1}{4}$ outside the summation to keep our workspace clean:

Here is the "Pro" move. You could expand $(k+1{)}^2$ into $k^2+2k+1$, but why do that? Let's use a substitution. Let $n=k+1$.

When $k=1$, $n=2$. When $k=10$, $n=11$. Our summation now becomes:

This is much cleaner! We are summing $n^2$ from $2$ to $11$. Since the standard formula for the sum of squares, ${\sum }_{i=1}^m{i}^2=\frac{m(m+1)(2m+1)}{6}$, starts from $1$, we can just calculate the sum from $1$ to $11$ and subtract the missing $1^2$ term.

The Final Victory

Let's calculate the sum from $1$ to $11$ using $m=11$:

Simplifying this, $12$ divided by $6$ is $2$, so we get $11\times 2\times 23=506$. Now, subtract the $1^2$ we didn't include: $506-1=505$.

Don't forget the $\frac{1}{4}$ waiting outside! Our total sum is $\frac{505}{4}$.

Finally, we equate this to the problem's given expression:

Solving for $A$ is just a matter of simple arithmetic:

And there you have it. We didn't get lost in the algebra; we simplified, we shifted, and we conquered. Remember, in physics and math, the most complex problems are often just simple concepts wearing a disguise. Keep your cool, simplify the general term, and the answer will reveal itself.