Let ⟨a_n⟩ be a sequence such that a_0 = 0, a_1 = 1/2 and 2a_{n+2} = 5a_{n+1} - 3a_n, n = 0, 1, 2, 3,

Q: Why did we rearrange the recurrence as $2(a_{n+2} - a_{n+1}) = 3(a_{n+1} - a_n)$ instead of using the characteristic equation method?

While the characteristic equation method is perfectly valid, the rearrangement method is often faster and more intuitive for linear recurrences where the sum of coefficients is zero. It immediately reveals the underlying geometric progression of the differences, which is a powerful tool in your JEE arsenal.

Q: I got $a_n = (3/2)^n - 1$. How do I know if this is correct for $a_0$ and $a_1$?

Always test your closed-form expression! If $n=0$, $a_0 = (3/2)^0 - 1 = 1 - 1 = 0$. If $n=1$, $a_1 = (3/2)^1 - 1 = 1.5 - 1 = 0.5$. Since both match the given initial conditions, you can proceed with absolute confidence.

Q: Can I just calculate the first few terms and look for a pattern instead of doing all this algebra?

For small values, yes, but in JEE Advanced, the pattern might be deceptive or too complex to generalize quickly. The algebraic derivation ensures you don't miss a subtle shift in the sequence behavior. Trust the math; it's the safest path to the correct answer.

Q: Why does the sum of $a_k$ result in an expression involving $a_{100}$?

The problem asks for the sum in terms of the sequence elements themselves. By finding the closed form of $a_n$, we essentially 'bridge' the gap between the summation and the individual terms. It's a beautiful way to simplify a long series into a compact, elegant result.

Q: Is there a risk of an off-by-one error when summing from $k=1$ to $100$?

Absolutely! Always double-check your summation limits. Since we are summing $a_k$ from $k=1$ to $100$, ensure your geometric series formula accounts for exactly 100 terms. Writing out the first and last terms of the sum is a great way to verify your indices.

Question

Let $\langle a_{n}\rangle$ be a sequence such that $a_{0}=0$, $a_{1}=\frac{1}{2}$ and $2a_{n+2}=5a_{n+1}-3a_{n}$, $n=0,1,2,3,...$ Then $\sum_{k=1}^{100}a_{k}$ is

Options:

(A) $3a_{99}-100$

(B) $3a_{100}-100$

(C) $3a_{99}+100$

(D) $3a_{100}+100$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to peel back the layers of a seemingly intimidating recurrence relation.

Our sequence $\langle a_n \rangle$ is defined by the relation:
$$2a_{n+2} = 5a_{n+1} - 3a_n$$
The starting points are given as $a_0 = 0$ and $a_1 = \frac{1}{2}$. At first glance, it looks like a standard linear recurrence, but there is a hidden elegance waiting to be discovered.

### The Art of Rearrangement

Whenever you see a recurrence where the coefficients sum to zero, pause. Look at the equation: $2a_{n+2} - 5a_{n+1} + 3a_n = 0$.

Since $2 - 5 + 3 = 0$, we can rewrite this as:
$$2a_{n+2} - 2a_{n+1} = 3a_{n+1} - 3a_n$$
By factoring, we obtain:
$$2(a_{n+2} - a_{n+1}) = 3(a_{n+1} - a_n)$$

Imagine you are standing on a staircase. If we define $d_n = a_{n+1} - a_n$, our equation becomes $2d_{n+1} = 3d_n$, or:
$$d_{n+1} = \frac{3}{2} d_n$$
We have just unmasked a **geometric progression**. The differences between consecutive terms scale by a factor of $\frac{3}{2}$ at every step.

### Telescoping: The Bridge to the General Term

We know $d_n$ is a geometric progression with the first term $d_0 = a_1 - a_0 = \frac{1}{2}$. Thus, its general term is:
$$d_k = \frac{1}{2} \left( \frac{3}{2} \right)^k$$

Think of $a_n$ as the sum of all the "jumps" taken to reach the $n$-th term:
$$a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k)$$
Since $a_0 = 0$, we sum the geometric progression:
$$a_n = \sum_{k=0}^{n-1} \frac{1}{2} \left( \frac{3}{2} \right)^k$$

Using the sum formula for a G.P., $S = a \frac{r^n - 1}{r - 1}$, we get:
$$a_n = \frac{1}{2} \cdot \frac{(\frac{3}{2})^n - 1}{\frac{3}{2} - 1}$$
The denominator $\frac{3}{2} - 1$ is exactly $\frac{1}{2}$, which cancels out perfectly. We are left with the elegant result:
$$a_n = \left( \frac{3}{2} \right)^n - 1$$

### The Final Summation

We need to calculate the sum $\sum_{k=1}^{100} a_k$. Substituting our expression for $a_k$, we have:
$$\sum_{k=1}^{100} \left( \left( \frac{3}{2} \right)^k - 1 \right) = \sum_{k=1}^{100} \left( \frac{3}{2} \right)^k - \sum_{k=1}^{100} 1$$

The second part is trivial: it is simply $100$. The first part is a geometric series sum:
$$\frac{\frac{3}{2} \left( (\frac{3}{2})^{100} - 1 \right)}{\frac{3}{2} - 1} = 3 \left( \left( \frac{3}{2} \right)^{100} - 1 \right)$$

Putting it all together, our total sum is:
$$3 \left( \left( \frac{3}{2} \right)^{100} - 1 \right) - 100$$
Since $a_{100} = (\frac{3}{2})^{100} - 1$, the final result can be expressed as:
**$$3a_{100} - 100$$**

This is the essence of JEE mathematics—not just calculation, but the discovery of structure. You have mastered the sequence; now go forth and master the exam.

Animated Solution for Mathematics - Sequence and Series: Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

Select Answer:

Visualized Solution

Analyzing the Given Recurrence

Rearranging the Recurrence

Defining the Difference Sequence

Identifying the Geometric Progression

Finding the General Term $d_{k}$

Expressing $a_{n}$ as a Telescoping Sum

Evaluating the Sum for $a_{n}$

Setting up the Final Summation

Calculating the G.P. Sum

Simplifying the Expression

Relating to $a_{100}$ and Final Answer

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Art of Rearrangement

Telescoping

The Final Summation

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