For m, n 0, let alpha(m, n) = integral from 0 to 2 of t^m (1 + 3t)^n dt. If 11*alpha(10, 6) + 18*a

Q: Wait, why did we choose $u = (1+3t)^n$ instead of $u = t^m$?

Great observation! If you chose $u = t^m$, you would get $du = mt^{m-1}dt$, which increases the complexity of the integral by keeping the $(1+3t)^n$ term intact. By choosing $u = (1+3t)^n$, we reduce the power of the binomial term, which is exactly what we need to create a recurrence relation that links $\alpha(m, n)$ to $\alpha(m+1, n-1)$.

Q: How do I know when to stop the integration by parts?

In problems involving recurrence relations, you don't 'solve' the integral to a single number immediately. You stop once you have successfully expressed the original integral in terms of another integral with different indices. The goal is to build a bridge between the two terms given in the problem statement.

Q: Is there a trick to evaluate the boundary term $[uv]_0^2$?

Always check your limits! At $t=0$, the term $t^{m+1}$ becomes $0$, which is a huge relief. At $t=2$, you simply substitute the value. The 'trick' is to keep your eyes on the target: $14^6$. Notice how $2 \times 7 = 14$. The boundary term $2^{m+1} 7^n$ is designed to lead you exactly there.

Q: Why does the coefficient $3n$ appear in the recurrence?

That comes from the Chain Rule! When you differentiate $u = (1+3t)^n$, you get $du = n(1+3t)^{n-1} \cdot \frac{d}{dt}(1+3t) \, dt$, which is $3n(1+3t)^{n-1} \, dt$. That factor of $3$ is crucial—without it, the coefficients in the final equation wouldn't align.

Q: Can I solve this using Beta functions?

While the integral looks like a Beta function, the limits $0$ to $2$ (instead of $0$ to $1$) make a direct Beta function substitution messy. Recurrence relations are the 'elegant' path here because they bypass the need to evaluate the integral explicitly, which is often the intended path in JEE Advanced.

Question

For $m, n > 0$, let $\alpha(m, n) = \int_0^2 t^m (1 + 3t)^n \, dt$. If $11\alpha(10, 6) + 18\alpha(11, 5) = p(14)^6$, then $p$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, complex integral: 
$$\alpha(m, n) = \int_0^2 t^m (1 + 3t)^n \, dt$$

It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for a hidden symmetry. Today, we are going to peel back that mask.

### The Strategy: Integration by Parts

When you see an integral with powers that look like they could 'trade' values—where one power could go up while the other goes down—your intuition should immediately scream **Integration by Parts**. We want to create a bridge between $\alpha(m, n)$ and $\alpha(m+1, n-1)$.

To do this, we set our stage. Let $u = (1 + 3t)^n$ and $dv = t^m \, dt$.

Because when we differentiate $u$, the power $n$ drops to $n-1$, and when we integrate $dv$, the power $m$ climbs to $m+1$, we are effectively shifting the 'weight' of the integral from one term to the other.

### The Calculus of Transformation

Applying the formula $\int u \, dv = [uv] - \int v \, du$, we get:

$$\alpha(m, n) = \left[ (1+3t)^n \cdot \frac{t^{m+1}}{m+1} \right]_0^2 - \int_0^2 \frac{t^{m+1}}{m+1} \cdot 3n(1+3t)^{n-1} \, dt$$

Look at that boundary term. At $t=0$, the $t^{m+1}$ term vanishes into nothingness.

At $t=2$, we get a beautiful constant: 
$$\frac{7^n \cdot 2^{m+1}}{m+1}$$

The integral term on the right is simply a scaled version of $\alpha(m+1, n-1)$. We have successfully built our bridge.

### The Recurrence Relation

By rearranging the terms, we arrive at the recurrence relation:

$$(m+1)\alpha(m, n) + 3n \alpha(m+1, n-1) = 2^{m+1} 7^n$$

This equation is the heartbeat of the problem. It tells us exactly how these integrals behave.

Now, the moment of truth: we substitute $m=10$ and $n=6$. The equation transforms into:

$$11\alpha(10, 6) + 18\alpha(11, 5) = 2^{11} 7^6$$

### The Final Reveal

We are almost there. The problem asks us to match this to $p(14)^6$. Look closely at $2^{11} 7^6$.

We can rewrite $2^{11}$ as $2^5 \cdot 2^6$. This gives us:

$$2^5 \cdot (2^6 \cdot 7^6) = 32 \cdot (2 \cdot 7)^6 = 32 \cdot (14)^6$$

Comparing this to $p(14)^6$, the value of **$p$** reveals itself as **$32$**.

### Why This Matters

This problem isn't just about finding a number. It is about learning to see the 'flow' of an integral.

By using Integration by Parts, we didn't just calculate; we transformed the problem into a simpler, more elegant state. Whenever you face a daunting integral in the exam, don't panic. Look for the recurrence and look for the bridge.

Animated Solution for Mathematics - Definite Integration: For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} (1 + 3 t)^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p (14)^{6}$ , then $p$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Integral $α (m, n)$

Choosing Integration by Parts

Differentiating $u$ and Integrating $d v$

Applying the Integration by Parts Formula

Evaluating the Boundary Terms

Identifying the Recurrence Relation

Simplifying the Recurrence Formula

Substituting $m = 10$ and $n = 6$

Solving for $p$

Final Conclusion & Takeaway

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Strategy

The Calculus of Transformation

The Recurrence Relation

The Final Reveal

Why This Matters

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For m,n>0, let α(m,n)=∫02​tm(1+3t)ndt. If 11α(10,6)+18α(11,5)=p(14)6, then p is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α(m,n)