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

Ratio and Proportion

There will be...

Question

There will be no term containing $x^{2 r}$ in the expansion of ${(x + x^{- 2})}^{n - 3}$ if $(n - 2 r)$ is positive but not a multiple of

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3$
$(x + \frac{1}{x^{2}})^{n - 3}$
$T_{r + 1} =^{n - 3} C_{r} x^{n - 3 - 3 r}$
For $x^{2 r}$
$n - 3 - 3 r = 2 r$
$n - 3 = 5 r$
or
$n - 2 r = 3 + 3 r$
$= 3 (1 + r)$
$= 3 k$
Hence $n - 2 r$ has to be a multiple of 3 for the existence of $x^{2 r}$

Suggest Corrections

Similar questions

If there is a term containing $x^{2 r} i n {(x + \frac{1}{x^{2}})}^{n - 3}$ , then :

x^{2 r}

{(x + \frac{1}{x^{2}})}^{n - 3}

(\frac{n - 2 r}{3})

x^{2 r}

{(x + \frac{1}{x^{2}})}^{n - 3}

(x + 1)^{n} - x^{n} - 1

x^{3} + x^{2} + x

1 + ω^{n} + ω^{2 n} = 3

n

(x + 1)^{n} - x^{n} - 1

x^{3} + x^{2} + x

n

3

3

(1 + ω^{n} + ω^{2 n}) = 0

Join BYJU'S Learning Program