Problem 6

Probability Level 4

The sum of the coefficients of all the integrals powers of $x$ in the expansion of ${ \left( 1+2\sqrt { x } \right) }^{ 40 }$ If the sum can be represented as $\frac { 1 }{ A } \left( { 3 }^{ B }+1 \right),$ where $A, B$ are positive integers less than 100. Find the value of $A+B+1$

The answer is 43.

1 solution

Surya Prakash
Aug 11, 2015

Let $C_{r} = ^{40}C_{r}$ .

The sum of coefficients of integral powers of $x$ in $(1+2\sqrt{x})^{40}$ is $C_{0} + 2^{2}C_2 + \ldots + 2^{40}C_{40}$

Let $f(x) = (1+x)^{40}$ .

So, The required sum can be written as $\dfrac{f(2)+f(-2)}{2}$

$=\dfrac{1}{2}(3^{40}+1)$

So, $A=2$ , $B=40$ and $C=1$ .

Therefore, $A+B+C = \boxed{43}$ .

Moderator note:

Simple standard approach with generating functions.

Problem 6

The answer is 43.

1 solution

Moderator note:

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