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Binomial Theorem question

2024 · 31 Jan · Shift 1 · Q52
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  5. /2024 · 31 Jan · Shift 1 · Q52

Binomial Theorem question

2024 · 31 Jan · Shift 1 · Q52

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
In the expansion of (1+x)(1−x2)(1+3x+3x2+1x3)5,xeq0(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x eq 0(1+x)(1−x2)(1+x3​+x23​+x31​)5,xeq0, the sum of the coefficients of x3x^3x3 and x−13x^{-13}x−13 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 118

  1. Simplify the given expression

We need the coefficients of x3x^3x3 and x−13x^{-13}x−13 in

(1+x)(1−x2)(1+3x+3x2+1x3)5.(1+x)(1-x^2)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5.(1+x)(1−x2)(1+x3​+x23​+x31​)5.

Notice that

1+3x+3x2+1x3=(1+1x)3.1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3} = \left(1+\frac{1}{x}\right)^3.1+x3​+x23​+x31​=(1+x1​)3.

Therefore,

(1+3x+3x2+1x3)5=(1+1x)15.\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 = \left(1+\frac{1}{x}\right)^{15}.(1+x3​+x23​+x31​)5=(1+x1​)15.

Also,

(1+x)(1−x2)=(1+x)2(1−x).(1+x)(1-x^2)=(1+x)^2(1-x).(1+x)(1−x2)=(1+x)2(1−x).

So the whole expression becomes

(1+x)2(1−x)(1+1x)15.(1+x)^2(1-x)\left(1+\frac{1}{x}\right)^{15}.(1+x)2(1−x)(1+x1​)15.

Now,

(1+1x)15=x−15(1+x)15.\left(1+\frac{1}{x}\right)^{15}=x^{-15}(1+x)^{15}.(1+x1​)15=x−15(1+x)15.

Hence the expression is

x−15(1+x)17(1−x).x^{-15}(1+x)^{17}(1-x).x−15(1+x)17(1−x).
  1. Rewrite in a more usable form

Let

F(x)=x−15(1+x)17(1−x).F(x)=x^{-15}(1+x)^{17}(1-x).F(x)=x−15(1+x)17(1−x).

Then

F(x)=x−15[(1+x)17−x(1+x)17].F(x)=x^{-15}\big[(1+x)^{17}-x(1+x)^{17}\big].F(x)=x−15[(1+x)17−x(1+x)17].

So coefficient extraction becomes easy.

If we want coefficient of xmx^mxm in F(x)F(x)F(x), it is the coefficient of xm+15x^{m+15}xm+15 in

(1+x)17−x(1+x)17.(1+x)^{17}-x(1+x)^{17}.(1+x)17−x(1+x)17.

That is,

[xm]F(x)=[xm+15](1+x)17−[xm+14](1+x)17.[x^m]F(x)=[x^{m+15}](1+x)^{17}-[x^{m+14}](1+x)^{17}.[xm]F(x)=[xm+15](1+x)17−[xm+14](1+x)17.

Using

[xr](1+x)17=(17r),[x^r](1+x)^{17}=\binom{17}{r},[xr](1+x)17=(r17​),

we get

[xm]F(x)=(17m+15)−(17m+14).[x^m]F(x)=\binom{17}{m+15}-\binom{17}{m+14}.[xm]F(x)=(m+1517​)−(m+1417​).
  1. Coefficient of x3x^3x3

Put m=3m=3m=3:

[x3]F(x)=(1718)−(1717)=0−1=−1.[x^3]F(x)=\binom{17}{18}-\binom{17}{17}=0-1=-1.[x3]F(x)=(1817​)−(1717​)=0−1=−1.

So coefficient of x3x^3x3 is

−1.-1.−1.
  1. Coefficient of x−13x^{-13}x−13

Put m=−13m=-13m=−13:

[x−13]F(x)=(172)−(171)=136−17=119.[x^{-13}]F(x)=\binom{17}{2}-\binom{17}{1}=136-17=119.[x−13]F(x)=(217​)−(117​)=136−17=119.

So coefficient of x−13x^{-13}x−13 is

119.119.119.
  1. Sum of the two coefficients

Therefore,

−1+119=118.-1+119=118.−1+119=118.

So the required integer is

118.\boxed{118}.118​.
  1. Comparison with stored answer

Stored correct answer = 118118118.

Our derived answer matches the stored answer.

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