Animated Solution for Mathematics - Sequence and Series: The sum, ${\sum }_{n=1}^7\frac{n(n+1)(2n+1)}{4}$ is equal to

Mastering Summation

Hello, future engineers! Today, we are going to look at a problem that, at first glance, might seem like a tedious arithmetic exercise. You see a summation, a product of terms, and a limit of seven.

Your instinct might be to start calculating for $n=1$, then $n=2$, and so on. But stop! In the world of JEE Advanced, we don't just calculate; we strategize. We look for the underlying structure.

Let's embark on this journey to solve the following expression with elegance and precision:

The Expansion Strategy

When you see a product of terms inside a summation, your brain should immediately scream: "Expand!" The expression $T_n=\frac{n(n+1)(2n+1)}{4}$ is a cubic polynomial waiting to be unleashed.

If we try to sum the product directly, we are fighting against the structure of the problem. Instead, let's simplify the numerator first.

First, we tackle the binomials: $(n+1)(2n+1)$. Using the distributive property, we get $2n^2+n+2n+1$, which simplifies to $2n^2+3n+1$.

Now, we bring that outer $n$ back into the fold. Multiplying $n$ by our quadratic gives us $2n^3+3n^2+n$.

Suddenly, our general term $T_n$ looks much more manageable:

We have transformed a product into a sum of powers. This is the first step of a master strategist.

The Power of Linearity

Now that we have our polynomial, we invoke one of the most beautiful properties in mathematics: the linearity of the summation operator. We are looking for $S={\sum }_{n=1}^7\frac{2n^3+3n^2+n}{4}$.

Because summation is linear, we can pull that constant $\frac{1}{4}$ completely outside the summation. Furthermore, we can distribute the summation across each term individually.

This transforms our single, intimidating problem into three distinct, standard problems:

Do you see what happened? We have broken the "monster" into three "pets." We know how to handle $\sum n$, $\sum n^2$, and $\sum n^3$. We are no longer guessing; we are executing a plan.

The Toolkit of Formulas

Now, we reach into our mathematical toolkit. For any upper limit $n=7$, we have the standard formulas ready to go:

1. The Sum of Natural Numbers: ${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$. For $n=7$, this is $\frac{7\times 8}{2}=28$.

2. The Sum of Squares: ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$. For $n=7$, this is $\frac{7\times 8\times 15}{6}=140$.

3. The Sum of Cubes: ${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$. Since we already know $\sum n=28$, this is simply $28^2=784$.

These formulas are the bedrock of series and sequences. Memorizing them is not just about passing an exam; it is about internalizing the patterns of nature.

The Final Synthesis

We are in the home stretch. We have our values: $784$, $140$, and $28$. Let's plug them back into our master equation:

Let's do the arithmetic with care. $2\times 784=1568$ and $3\times 140=420$. Adding these to $28$ gives us $1568+420+28=2016$.

Finally, we divide by $4$:

And there it is! The result is 504.

Look at what we achieved. We didn't just crunch numbers; we decomposed a complex expression, applied the linearity of operators, utilized standard power-sum formulas, and synthesized the result.

This is the essence of JEE Advanced problem-solving. It is not about brute force; it is about clarity, structure, and the confidence to break down any problem, no matter how intimidating it looks at first. Keep practicing these steps, and soon, you won't just be solving problems—you'll be mastering them.