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Binomial Theorem question

2016 · 9 Apr · Shift 1 · Q24
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  5. /2016 · 9 Apr · Shift 1 · Q24

Binomial Theorem question

2016 · 9 Apr · Shift 1 · Q24

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
For x ∈\in∈ R, x eee -1, if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + . . . . + x2016 = ∑i=02016ai xi,  \sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,i=0∑2016​ai​xi, then a17 is equal to :
  1. A
    2017!17!   2000!{{2017!} \over {17!\,\,\,2000!}}17!2000!2017!​
  2. B
    2016!17!   1999!{{2016!} \over {17!\,\,\,1999!}}17!1999!2016!​
  3. C
    2017!2000!{{2017!} \over {2000!}}2000!2017!​
  4. D
    2016!16!{{2016!} \over {16!}}16!2016!​
View written solutionFree

Correct answer: A

  1. Interpret the given sum

The expression is

(1+x)2016+x(1+x)2015+x2(1+x)2014+⋯+x2016.(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\cdots + x^{2016}.(1+x)2016+x(1+x)2015+x2(1+x)2014+⋯+x2016.

Its general term is

xk(1+x)2016−k,k=0,1,2,…,2016.x^k(1+x)^{2016-k}, \qquad k=0,1,2,\dots,2016.xk(1+x)2016−k,k=0,1,2,…,2016.

So the whole sum is

S=∑k=02016xk(1+x)2016−k.S=\sum_{k=0}^{2016} x^k(1+x)^{2016-k}.S=k=0∑2016​xk(1+x)2016−k.

We are given

S=∑i=02016aixi,S=\sum_{i=0}^{2016} a_i x^i,S=i=0∑2016​ai​xi,

and we need to find a17a_{17}a17​.

  1. Recognize it as a geometric-type sum

Let

a=1+x,b=x.a=1+x, \qquad b=x.a=1+x,b=x.

Then

S=a2016+ba2015+b2a2014+⋯+b2016.S=a^{2016}+ba^{2015}+b^2a^{2014}+\cdots+b^{2016}.S=a2016+ba2015+b2a2014+⋯+b2016.

This is the standard identity

an+an−1b+⋯+bn=an+1−bn+1a−b,a≠b.a^n+a^{n-1}b+\cdots+b^n=\frac{a^{n+1}-b^{n+1}}{a-b}, \qquad a\ne b.an+an−1b+⋯+bn=a−ban+1−bn+1​,a=b.

Here a−b=(1+x)−x=1a-b=(1+x)-x=1a−b=(1+x)−x=1, so

S=(1+x)2017−x2017.S=(1+x)^{2017}-x^{2017}.S=(1+x)2017−x2017.
  1. Expand and identify the coefficient of x17x^{17}x17

Now,

(1+x)2017=∑r=02017(2017r)xr.(1+x)^{2017}=\sum_{r=0}^{2017} \binom{2017}{r}x^r.(1+x)2017=r=0∑2017​(r2017​)xr.

Hence

S=(1+x)2017−x2017=∑r=02017(2017r)xr−x2017.S=(1+x)^{2017}-x^{2017} =\sum_{r=0}^{2017} \binom{2017}{r}x^r - x^{2017}.S=(1+x)2017−x2017=r=0∑2017​(r2017​)xr−x2017.

Since we need the coefficient of x17x^{17}x17 and 17≠201717\ne 201717=2017, the term −x2017-x^{2017}−x2017 does not affect it.

Therefore,

a17=(201717).a_{17}=\binom{2017}{17}.a17​=(172017​).
  1. Write in factorial form
a17=(201717)=2017!17! 2000!.a_{17}=\binom{2017}{17}=\frac{2017!}{17!\,2000!}.a17​=(172017​)=17!2000!2017!​.
  1. Compare with options

Option A is

2017!17! 2000!,\frac{2017!}{17!\,2000!},17!2000!2017!​,

which matches exactly.

So the correct option is A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.

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