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  • Please help! - Binomial theorem and its simple applications - JEE Main

The coefficient of x^{7} in the expansion of  (1-x-x^{2}+x^{3})^{6}  is

  • Option 1)

    –144

  • Option 2)

    132

  • Option 3)

    144

  • Option 4)

    –132

 

Answers (1)

best_answer

As we learnt in

General Term in the expansion of (x+a)^n -

T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}
 

- wherein

Where r\geqslant 0 \, and \, r\leqslant n

r= 0,1,2,----n

 

 We have to convert (1 - x - x2 + x3)6

Into normal expansion 

We get \left[(1-x)-x^{2}(1-x) \right ]^{6}

\Rightarrow\ \; \left[(1-x)(1-x^{2}) \right ]^{6}

\Rightarrow\ \; (1-x)^{6}(1-x^{2})^{6}

For coefficient of x7 in (1-x)^{6}(1-x^{2})^{6}

\Rightarrow\ \; \left(1-\ ^{6}C_{1}x+\ ^{6}C_{2}x^{2}.................. \right )\left(1-\ ^{6}C_{1}x^{2}+\ ^{6}C_{2}x^{4}-^{6}C_{3}x^{6} \right )

\Rightarrow\ \; ^{6}C_{1}.\ ^{6}C_{3}-^{6}C_{3}.\ ^{6}C_{2}+^{6}C_{5}.\ ^{6}C_{1}

\Rightarrow\ \; 6\times 20-20\times 15+36

\Rightarrow\ \; 120-300+36=-144

Correct option is 3.

 

 

 


Option 1)

–144

This is the correct option.

Option 2)

132

This is an incorrect option.

Option 3)

144

This is an incorrect option.

Option 4)

–132

This is an incorrect option.

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Aadil

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