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  • Confused! kindly explain, - Binomial theorem and its simple applications - JEE Main

  The term independent of x in the binomial expansion of

\left ( 1-\frac{1}{x}+3x^{5} \right )\left ( 2x^{2}-\frac{1}{x} \right )^{8}is:

  • Option 1)

    400

  • Option 2)

    496

  • Option 3)

    -400

  • Option 4)

    -496

 

Answers (1)

best_answer

As we learnt in

Independent term -

A term which does not contains x, then 

\dpi{120} x^{\left ( n,r \right )}= x^{0}

\therefore \left ( n,r \right )= 0 find r

 

- wherein

Where r\geqslant 0

 

 \left(1-\frac{1}{x}+3x^{5} \right )\left(2x^{2}-\frac{1}{x} \right )^{8}

General term of \left(2x^{2}-\frac{1}{x} \right )^{8} is ^{8}C_{r}(2x^{2})^{8-r} \left(\frac{-1}{x} \right )^{r}

\Rightarrow\ \; ^{8}C_{r}\ 2^{8-r},\ x^{16-2r-r}\ x(-1)^{r}

\Rightarrow\ \; ^{8}C_{r}\ 2^{8-r},\ x^{16-3r}\ x(-1)^{r}

We need the powers x0, x1, x-5 

 16-3r=0\ \;\Rightarrow\ \; r=\frac{16}{3}, inot possible.

16-3r=1\ \;\Rightarrow\ \; r=5,

For n = 5, we get ^{8}C_{5}\ 2^{3} (-1)^{5}=\frac{-8\times 8\times 7\times 6}{6}=-448

If 16 - 3r = -5 

r = 7, we get ^{8}C_{7}\ 2^{1}. (-1)^{1}=-16

SQ we get  \left(1-\frac{1}{x}+3x^{5} \right )\left(-448x-16^{-5} \right )=448-48=400

Correct option is 1.

 

 


Option 1)

400

This is the correct option.

Option 2)

496

This is an incorrect option.

Option 3)

-400

This is an incorrect option.

Option 4)

-496

This is an incorrect option.

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Plabita

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