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Binomial Theorem question

2020 · 7 Jan · Shift 2 · Q22
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  5. /2020 · 7 Jan · Shift 2 · Q22

Binomial Theorem question

2020 · 7 Jan · Shift 2 · Q22

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The coefficient of x7 in the expression (1 + x)10 + x(1 + x)9 + x2(1 + x)8 + ......+ x10 is:
  1. A
    120
  2. B
    330
  3. C
    420
  4. D
    210
View written solutionFree

Correct answer: B

  1. Write the given expression as a sum

The expression is

(1+x)10+x(1+x)9+x2(1+x)8+⋯+x10.(1+x)^{10}+x(1+x)^9+x^2(1+x)^8+\cdots+x^{10}.(1+x)10+x(1+x)9+x2(1+x)8+⋯+x10.

This is

∑k=010xk(1+x)10−k.\sum_{k=0}^{10} x^k(1+x)^{10-k}.k=0∑10​xk(1+x)10−k.
  1. Factor out the common term

Observe that

xk(1+x)10−k=(1+x)10(x1+x)k.x^k(1+x)^{10-k}=(1+x)^{10}\left(\frac{x}{1+x}\right)^k.xk(1+x)10−k=(1+x)10(1+xx​)k.

So,

∑k=010xk(1+x)10−k=(1+x)10∑k=010(x1+x)k.\sum_{k=0}^{10} x^k(1+x)^{10-k} =(1+x)^{10}\sum_{k=0}^{10}\left(\frac{x}{1+x}\right)^k.k=0∑10​xk(1+x)10−k=(1+x)10k=0∑10​(1+xx​)k.

This is a finite geometric series with ratio

r=x1+x.r=\frac{x}{1+x}.r=1+xx​.

Hence,

∑k=010(x1+x)k=1−r111−r.\sum_{k=0}^{10}\left(\frac{x}{1+x}\right)^k =\frac{1-r^{11}}{1-r}.k=0∑10​(1+xx​)k=1−r1−r11​.

Now,

1−r=1−x1+x=11+x.1-r=1-\frac{x}{1+x}=\frac{1}{1+x}.1−r=1−1+xx​=1+x1​.

Therefore,

(1+x)10⋅1−r111−r=(1+x)10⋅(1+x)(1−(x1+x)11).(1+x)^{10}\cdot \frac{1-r^{11}}{1-r} =(1+x)^{10}\cdot (1+x)\left(1-\left(\frac{x}{1+x}\right)^{11}\right).(1+x)10⋅1−r1−r11​=(1+x)10⋅(1+x)(1−(1+xx​)11).

So the whole expression becomes

(1+x)11−x11.(1+x)^{11}-x^{11}.(1+x)11−x11.
  1. Find the coefficient of x7x^7x7

Since

(1+x)11−x11,(1+x)^{11}-x^{11},(1+x)11−x11,

the term −x11-x^{11}−x11 does not affect the coefficient of x7x^7x7.

Thus the required coefficient is the coefficient of x7x^7x7 in (1+x)11(1+x)^{11}(1+x)11, which is

(117)=(114)=330.\binom{11}{7}=\binom{11}{4}=330.(711​)=(411​)=330.
  1. Check options
  • A: 120120120
  • B: 330330330 ✅
  • C: 420420420
  • D: 210210210

So the correct option is B.

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