JEE Advanced Binomial Theorem 2

Calculus Level 5

$1+\left ( \dfrac{1}{3}\times\dfrac{1}{2} \right ) + \left ( \dfrac{1}{3}\times\dfrac{4}{3}\times\dfrac{1}{2^2}\times\dfrac{1}{2!} \right )+\left ( \dfrac{1}{3}\times\dfrac{4}{3}\times\dfrac{7}{3}\times\dfrac{1}{2^3}\times\dfrac{1}{3!} \right ) +\cdots n \text{ terms}$

For natural numbers $n$ , let $S_n$ be the value of the above expression

Suppose that for positive integers $a,b,c,d$ with $\text{gcd}(a,b) = \text{gcd}(c,d) = 1$ , we have

$\displaystyle\lim_{n\rightarrow\infty} S_n=\left (\dfrac{a}{b}\right )^\dfrac{c}{d}$

What is the value of $a+b+c+d$ ?

This is a part of My Picks for JEE Advanced 2

The answer is 7.

1 solution

Kishlaya Jaiswal
Feb 25, 2015

We observe the expansion of $(1-x)^{-(n+1)}$ .using Negative Binomial Theorem (aka Newton's Generalized Binomial Theorem)

$\begin{aligned} \left(\frac{1}{1-x}\right)^{n+1} & = & \sum_{k \geq 0} \binom{n+k}{n} x^k \\ & = & \binom{n}{n} + \binom{n+1}{n} x + \binom{n+2}{n} x^2 + \binom{n+3}{n} x^3 + \ldots \\ & = & 1 + (n+1)x + \frac{(n+1)(n+2)}{2!} x^2 + \frac{(n+1)(n+2)(n+3)}{3!} x^3 + \ldots \\ \end{aligned}$

Plugging $x=\frac{1}{2}$ and $n=\frac{-2}{3}$ , yields our desired result as

$2^{1/3} = \left(\frac{1}{1-\frac{1}{2}}\right)^{1/3} = 1 + \frac{1}{3}\times \frac{1}{2} + \frac{1}{3}\times \frac{4}{3}\times \frac{1}{2^2}\times \frac{1}{2!} \\ \qquad + \frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}\times \frac{1}{2^3}\times \frac{1}{3!} + \ldots$

Thus, $\left(\frac{a}{b}\right)^{\frac{c}{d}} = \left(\frac{2}{1}\right)^{\frac{1}{3}}$

And hence, $a+b+c+d = 2+1+1+3 = \boxed{7}$

Same way! except yours is much better, i couldnt observe it so well so i found a recurrence between the coefficient of powers of $\frac{1}{6^{r-1}}$

namely $\\ { a }_{ i+1 }(i+2)={ a }_{ i }(3i+4)\\$

(i greater than or 0, also $a_i$ = $T_{r-2})$ where t is term of series (and (r-2) is its subscript)

and then used generating function to get a differential which upon solving gave me

$-(\frac { 1 }{ (3x-1)^{ \frac { 1 }{ 3 } } } +1)\frac { 1 }{ x } \quad =\quad { T }_{ 2 }+x{ T }_{ 3 }+{ x }^{ 2 }{ T }_{ 4 }....$

now i put 1/6 = x , then again multiplied by 1/6 and added $t_1$ =1

to get $2^{1/3}$

thus 7,

Awesome question @Pranjal , thanks cause i learnt the whole of generating functions to solve this problem and it was worth it,

Mvs Saketh - 6 years, 3 months ago

Yep, generating functions is also a nice way to approach it.

Actually, I never thought of proceeding with it.

Nice Job. +1

Kishlaya Jaiswal - 6 years, 3 months ago

How to use generating function for obtaining what is described as namely.

Athul Nambolan - 6 years, 2 months ago

Why don't we directly use $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ...$ with $n = -1/3$ and $x = -1/2$ instead of using $(1-x)^{-(n+1)}$ ?

Siddhartha Srivastava - 6 years, 3 months ago

Just because the expansion of $(1-x)^{-(n+1)}$ looks more cleaner than the other one, to me.

$(1-x)^{-(n+1)} = \sum_{k=0}^\infty {n+k \choose n} x^k$

Anyway, you can also proceed using $(1+x)^{n} = 1+nx+\frac{n(n-1)}{2!}x^2 + \ldots$ . And actually, there is no difference between both of them because eventually you are taking $n = \frac{-1}{3}$ (which is negative) whereas the expansion which I have used is infact, the general expression for negative $n$ . So actually, you can say that they both are one and the same thing.

Kishlaya Jaiswal - 6 years, 3 months ago

To each his own I guess. I was just asking because I thought the formula I wrote is more common. NCERT jindabad. xD

Siddhartha Srivastava - 6 years, 3 months ago

How can we prove that the limit Sn (n tends to infinity) turns out to be that binomial term (1-x)^p where x=1/2 & p =-1/3

Mayank Jha - 4 years, 2 months ago

JEE Advanced Binomial Theorem 2

This is a part of My Picks for JEE Advanced 2

The answer is 7.

1 solution

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