GATE CSE 2016 Set 1 | Question: 26

Question

The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.

Comments

Very good method @Sourav.Sharma

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The problem can read as :

You are given 3 buckets and you need to pick out 12 items, but each bucket only allows you to pick at least 3 items in one go (not less than that)

We can easily shift the origin from 3 to 0 (as 3 items are always compulsory to pick). So 9 will be picked anyway. We can only decide on the remaining 3.

And now we only need to pick 3 items (and we are allowed to pick any number of items starting from 1, 2,)

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Answer 1

(x^3+x^4+x^5+x^6+- - -)^3

(x^3(1+x+x^2+x^3+- - -))^3

x^9(1+x+x^2+x^3+ - - -)^3

[note : (1+x+x^2+x^3+ - - -)

 sum of infinite no: 1 / 1-r

r = x ]

==>x^9(1/(1-x)^3) = x^9((1-x)^-3)

==> x^9* 3-1+rCr            [ By n-1+rCr]

=> x^9*x^3(3-1+3C3)    [x^12 = x^9*x^3]

==> x^12(5C3)

==> 10x^12

Ans: 10

Answer 2

 

 

I go by this approach, you can tell me yours.

This is very basic.

 

Method 2: extended binomial theorem

Comments

really love your first approach…..thank u

Answer 3

We have been given the sequence, $(x^3+x^4+x^5+x^6+...)^3$.

Start by taking $x^3$ common:

$(x^3+x^4+x^5+x^6+...)^3=\left [ x^3(1+x+x^2+x^3+...) \right ]^3$

This can now be written as follows by distributing the power:

$(x^3+x^4+x^5+x^6+...)^3=x^9*(1+x+x^2+x^3+...)^3$

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Notice that:

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

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Thus, we now have:

$x^9*(1+x+x^2+x^3+...)^3=x^9*\frac{1}{(1-x)^3}$

Clearly, the coefficient of $x^{12}$ will be the coefficient of $x^3$ in the expansion of $\frac{1}{(1-x)^3}$. The goal now is to find this expansion.

Start with:

$\frac{1}{(1-x)} = 1+x+x^2+...+x^5+...$

Differentiate both sides with respect to $x$:

$\frac{1}{(1-x)^2} = 1+2x+...+5x^4+...$

Differentiate both sides with respect to $x$ againt:

$\frac{2}{(1-x)^3} = 2+...+20x^3+...$

Distribute the $2$ in numerator to the RHS:

$\frac{1}{(1-x)^3} = 1+...+10x^3+...$

Now, when you multiply this with $x^9$, the coeffiicient of $x^{12}$ will come out as $10$.

ANSWER: 10.

Answer 4

$\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$