Animated Solution for Mathematics - Binomial Theorem: Let $(2x^2+3x+4{)}^{10}={\sum }_{r=0}^{20}{a}_r{x}^r$. Then $\frac{a_7}{a_{13}}$ is equal to

Imagine standing before the expression $(2x^2+3x+4{)}^{10}$. It looks like a daunting, impenetrable wall of algebra. Most students would immediately panic, wondering how to expand such a massive expression.

But as an elite JEE aspirant, you don't see a wall; you see a puzzle waiting to be solved. We are tasked with finding the ratio $\frac{a_7}{a_{13}}$, where $a_7$ and $a_{13}$ are coefficients of $x^7$ and $x^{13}$ respectively.

The Big Brother of Binomial

When we face a trinomial, the standard Binomial Theorem is not enough. We need the Multinomial Theorem. For any expression $(x+y+z{)}^n$, the general term is given by:

This is subject to the constraint $r_1+r_2+r_3=n$. In our case, $n=10$, and our terms are $2x^2$, $3x$, and $4$.

Substituting these, our general term becomes:

Simplifying this, we group the constants and the $x$ terms:

This is the heartbeat of the problem. The power of $x$ is $2r_1+r_2$.

To find $a_7$, we need the exponent of $x$ to be $7$. This gives us the system: $2r_1+r_2=7$ and $r_1+r_2+r_3=10$.

By testing non-negative integer values for $r_1$, we find the valid triplets $(r_1,r_2,r_3)$: $(0,7,3)$, $(1,5,4)$, $(2,3,5)$, and $(3,1,6)$. Each triplet contributes a term to the coefficient $a_7$.

We write $a_7$ as the sum of these four terms. Do not calculate them! The secret to JEE success is recognizing when to hold back on arithmetic.

Now, we repeat the process for $a_{13}$. We set $2r_1+r_2=13$ and $r_1+r_2+r_3=10$.

Testing values for $r_1$, we find the triplets: $(3,7,0)$, $(4,5,1)$, $(5,3,2)$, and $(6,1,3)$. Notice the elegance here? The values of $r_2$ are identical to those in $a_7$.

This is not a coincidence; it is the mathematical structure revealing itself.

When we form the ratio $\frac{a_7}{a_{13}}$, we compare the terms. Because $4=2^2$, every term in $a_7$ is exactly $8$ times the corresponding term in $a_{13}$.

The multinomial coefficients $\frac{10!}{r_1!r_2!r_3!}$ cancel out, the powers of $3$ cancel out, and we are left with a beautiful, clean $8$.

A Wizard's Shortcut

For those who want to master the exam, consider the symmetry method. Define $P(x)=(2x^2+3x+4{)}^{10}$.

If you evaluate $x^{20}P(\frac{2}{x})$, you get $(8+6x+4x^2{)}^{10}$, which is $2^{10}P(x)$. By equating the coefficients of $x^{13}$ on both sides, the ratio $\frac{a_7}{a_{13}}$ emerges instantly as $2^3=8$.

This is the power of perspective. Whether you use the brute-force (but logical) triplet method or the elegant symmetry trick, the final answer is 8.