Let {a_n} from n=0 to ∞ be a sequence such that a₀ = a₁ = 0 and a_{n+2} = 2a_{n+1} - a_n + 1 for all

Q: Wait, why do we define $b_n = a_{n+1} - a_n$? Is there a reason for this?

Great question! In recurrence relations, when you see $a_{n+2} - 2a_{n+1} + a_n = 1$, it is a classic signal to look at the 'first differences'. By defining $b_n$ as the difference between consecutive terms, we reduce the order of the recurrence. It turns a complex-looking relation into a simple linear one, $b_{n+1} = b_n + 1$, which is much easier to solve.

Q: Why do we differentiate the power series twice?

We differentiate because our expression for $a_n$ is a quadratic, specifically $a_n = \frac{n(n-1)}{2}$. The power series for $\sum x^n$ gives us terms of $x^n$. Differentiating once brings down an $n$, and differentiating a second time brings down the $(n-1)$. This perfectly matches the $n(n-1)$ structure we need for our sum!

Q: Can we use the method of undetermined coefficients instead of power series?

Yes, you absolutely can! You could assume the solution for $a_n$ is of the form $An^2 + Bn + C$ and solve for the constants using the initial conditions $a_0=0$ and $a_1=0$. However, the power series method is often faster for infinite sums because it directly gives you the closed-form result for the summation without needing to find the general term $a_n$ first.

Question

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0 = a_1 = 0$ and $a_{n+2} = 2a_{n+1} - a_n + 1$ for all $n \geq 0$. Then, $\sum_{n=2}^{\infty} \frac{a_n}{7^n}$ is equal to

Options:

(A) $\frac{6}{343}$

(B) $\frac{7}{216}$

(C) $\frac{8}{343}$

(D) $\frac{49}{216}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering the hidden architecture of a sequence.

We are given a recurrence relation: 
$$a_{n+2} = 2a_{n+1} - a_n + 1$$ 
with the starting conditions $a_0 = 0$ and $a_1 = 0$.

Our mission is to find the value of the infinite sum:
$$S = \sum_{n=2}^{\infty} \frac{a_n}{7^n}$$

At first glance, this looks like a daunting task. The secret lies in seeing the 'soul' of the sequence.

### Decoding the Recurrence

Let us look at the recurrence relation again: $a_{n+2} = 2a_{n+1} - a_n + 1$. If we rearrange this, we get:
$$a_{n+2} - a_{n+1} = a_{n+1} - a_n + 1$$

This is the 'Aha!' moment. If we define a new sequence $b_n = a_{n+1} - a_n$, the equation becomes $b_{n+1} = b_n + 1$.

This tells us that the difference between consecutive terms is increasing by exactly $1$ at every step. Since $b_0 = a_1 - a_0 = 0 - 0 = 0$, the sequence $b_n$ is simply the sequence of natural numbers: $b_n = n$.

### The Quadratic Nature

Now that we know the differences, we can reconstruct $a_n$. We know that:
$$a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k)$$

Since $a_0 = 0$ and $a_{k+1} - a_k = b_k = k$, we have:
$$a_n = \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2}$$

Our sequence is quadratic. If you were to plot these points, you would see a perfect parabola emerging from the discrete values.

### The Power Series Magic

Now, we tackle the infinite sum:
$$S = \sum_{n=2}^{\infty} \frac{a_n}{7^n} = \sum_{n=2}^{\infty} \frac{n(n-1)}{2 \cdot 7^n}$$

We can pull the constant $1/2$ out:
$$S = \frac{1}{2} \sum_{n=2}^{\infty} n(n-1) \left(\frac{1}{7}ight)^n$$

This is where the beauty of calculus shines. We start with the geometric series:
$$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$

Differentiating once gives:
$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$

Differentiating a second time gives:
$$\sum_{n=2}^{\infty} n(n-1) x^{n-2} = \frac{2}{(1-x)^3}$$

To match our target sum, we multiply by $x^2$ to get:
$$\sum_{n=2}^{\infty} n(n-1) x^n = \frac{2x^2}{(1-x)^3}$$

### The Grand Finale

We are almost there. We substitute $x = 1/7$ into our derived formula. Our sum $S$ is:
$$S = \frac{1}{2} 	imes \frac{2x^2}{(1-x)^3} = \frac{x^2}{(1-x)^3}$$

Plugging in $x = 1/7$: the numerator is $(1/7)^2 = 1/49$, and the denominator is:
$$(1 - 1/7)^3 = (6/7)^3 = \frac{216}{343}$$

Thus, the final calculation yields:
$$S = \frac{1/49}{216/343} = \frac{1}{49} 	imes \frac{343}{216} = \mathbf{\frac{7}{216}}$$

The elegance of this result—the way the powers of $7$ cancel out—is the hallmark of a well-crafted JEE problem. You have successfully navigated the recurrence, the summation, and the power series.

Animated Solution for Mathematics - Sequence and Series: Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

Select Answer:

Visualized Solution

Understanding the Recurrence

Rearranging the Recurrence

General Term of $b_{n}$

Finding $a_{n}$ via Telescoping Sum

Calculating $a_{n}$

Setting up the Infinite Sum

The Power Series Tool

First Derivative

Second Derivative

Matching the Powers

Substituting $x = 1/7$

Final Calculation

The Final Answer

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Decoding the Recurrence

The Quadratic Nature

The Power Series Magic

The Grand Finale

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