CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Special Integrals - 2

Show that rat...

Question

Show that ratio of the coefficient of x¹⁰ in (1-x²)¹⁰ and the term independent of x in (x-2/x)¹⁰ is 1 :32 .

Open in App

Solution

Dear student

$We know that general term in the expansion of {(a - b)}^{n} is T_{r + 1} = {(- 1)}^{r} . {}^{n}C_{r} . a^{n - r} . b^{r} So, general term in the expansion of {(1 - x^{2})}^{10} is T_{r + 1} = {(- 1)}^{r}^{10} C_{r} {(1)}^{10 - r} . {(x^{2})}^{r} T_{r + 1} =^{10} C_{r} {(- 1)}^{r} {(x^{2})}^{r} Putting r = 5, we get T_{6} = {(- 1)}^{5} .^{10} C_{5} {(x^{2})}^{5} = -^{10} C_{5} . x^{10} So, Coefficient of x^{10} = -^{10} C_{5} general term in the expansion of {(x - \frac{2}{x})}^{10} is T_{r + 1} = {(- 1)}^{r}^{10} C_{r} {(x)}^{10 - r} \times {(\frac{2}{x})}^{r} T_{r + 1} = {(- 1)}^{r}^{10} C_{r} . 2^{r} . {(x)}^{10 - 2 r} Putting 10 - 2 r = 0, we get r = 5 So, term independent of x : T_{6} = {(- 1)}^{5}^{10} C_{5} . 2^{5} = - 32^{10} C_{5} Hence their ratio = (-^{10} C_{5}) : (- 32^{10} C_{5}) = 1 : 32$

Regards

Suggest Corrections

thumbs-up

0

Similar questions

Q. Prove that the ratio of the coefficient of

x^{10}

(1 - x^{2})^{10}

and the term independent of x in

{(x - \frac{2}{x})}^{10}

Q. The ratio of the coefficient of

x^{10}

(1 - x^{2})^{10}

and the term independent of

x

{(x - \frac{2}{x})}^{10}

Q. The ratio of the coefficients of

x^{10}

(1 - x^{2})^{10}

and the term independent of

x

{(x - \frac{2}{x})}^{10}

Q. The term independent of

x

is in the expansion of

{(1 - x)}^{2}

{(x + \frac{1}{x})}^{10}

Q. The term independent of

x

in the expansion of

(1 - x)^{2} {(x + \frac{1}{x})}^{10}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Special Integrals - 2

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Special Integrals - 2

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon