The sum, sum_{n=1}^{7} (n(n+1)(2n+1))/4 is equal to.

Q: Why can't I just calculate each term from $n=1$ to $7$ and add them up?

You absolutely can, and for $n=7$, it is a valid strategy! However, the power of this method lies in its scalability. If the problem asked for the sum up to $n=100$, calculating term-by-term would be impossible. Mastering the formulaic approach is a crucial skill for JEE Advanced.

Q: Is there a shortcut for $\sum \frac{n(n+1)(2n+1)}{4}$?

Yes! Notice that $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. The expression inside your sum is $\frac{n(n+1)(2n+1)}{4}$, which is $\frac{6}{4} \times \frac{n(n+1)(2n+1)}{6} = 1.5 \times n^2$. So, the sum is simply $1.5 \times \sum_{n=1}^7 n^2$. This is a much faster way to reach the answer!

Q: Do I really need to memorize $\sum n^3 = (\sum n)^2$?

Yes, it is a beautiful identity known as Nicomachus's Theorem. It is not just a memory aid; it represents a deep geometric connection between the sum of cubes and the square of the sum of natural numbers. It will save you significant time in exams.

Question

The sum, $\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to _________ .

Sigma Percentile · Accepted Answer

### The Art of Summation: A Journey Through Series

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a math problem; we are uncovering the elegance hidden within a finite series.

When you look at the expression

$$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$$

it is easy to feel overwhelmed by the product of terms. But remember, in physics and mathematics, complexity is often just a mask for a simpler, more beautiful structure waiting to be revealed. Let us peel back that mask together.

### The Power of Linearity

Our first step is to simplify the landscape. We are dealing with a summation, and summation is a linear operator. This means it respects addition and scalar multiplication.

We can pull that constant $\frac{1}{4}$ right out of the sum, leaving us with:

$$\frac{1}{4} \sum_{n=1}^{7} n(n+1)(2n+1)$$

By doing this, we have isolated the core of the problem. We are no longer looking at a fraction; we are looking at a polynomial sum, which is much friendlier to handle.

### Algebraic Expansion: The Bookkeeping of Math

Now, we must expand the product $n(n+1)(2n+1)$. Think of algebra as the bookkeeping of mathematics. We distribute the terms carefully.

First, $n(n+1)$ gives us $n^2 + n$. Then, we multiply this by $(2n+1)$.

Distributing $n^2$ into $(2n+1)$ gives $2n^3 + n^2$, and distributing $n$ into $(2n+1)$ gives $2n^2 + n$. Combining these, we get the cubic polynomial:

$$2n^3 + 3n^2 + n$$

This is the heart of our series. We have transformed a product of three factors into a sum of three simple power terms.

### The Toolbox: Standard Summation Formulas

To solve this, we need our tools. In the JEE arsenal, you must have the fundamental summation formulas memorized:

$$\sum n = \frac{n(n+1)}{2}$$

$$\sum n^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum n^3 = \left( \frac{n(n+1)}{2} \right)^2$$

These are not just formulas; they are the keys to unlocking series problems. For $n=7$, we calculate each:

$$\sum_{n=1}^{7} n = 28, \quad \sum_{n=1}^{7} n^2 = 140, \quad \sum_{n=1}^{7} n^3 = 784$$

Notice how the sum of cubes is just the square of the sum of $n$? That is the beauty of Nicomachus's Theorem in action.

### The Final Assembly

We are now at the final stage. We substitute our values back into the expression:

$$S = \frac{1}{4} [2(784) + 3(140) + 28]$$

Performing the arithmetic, we get:

$$S = \frac{1}{4} [1568 + 420 + 28] = \frac{2016}{4} = 504$$

The journey is complete. We have navigated the expansion, applied the formulas, and arrived at the solution.

Remember, every complex problem is just a series of simple steps waiting for you to take them. The final answer is **504**. Keep practicing, keep questioning, and most importantly, keep falling in love with the process!

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

Enter Numerical Value:

Visualized Solution

Analyzing the Summation Expression

Applying the Linearity Property

Expanding the Polynomial

Simplifying the Cubic Expression

Distributing the Summation Operator

Standard Summation Formulas

Evaluating $\sum n$ for $n = 7$

Evaluating $\sum n^{2}$ for $n = 7$

Evaluating $\sum n^{3}$ for $n = 7$

Substituting the Calculated Sums

Calculating the Final Sum

Final Division

The Sigma Insight: Sum of Special Series

The Art of Summation

The Power of Linearity

Algebraic Expansion

The Toolbox

The Final Assembly

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