If $(1 + x)^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ then

a_{n + 1} >

a_{n}

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

a_{n + 1} <

a_{n}

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

a_{n - 3} = a_{n + 3}

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

none of these

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

Open in App

Solution

The correct options are
A $a_{n - 3} = a_{n + 3}$
B $a_{n + 1} <$ $a_{n}$
In the above given binomial expansion
$a_{r} =^{2 n} C_{r}$
Therefore
$a_{n + 1} =^{2 n} C_{n + 1}$
And
$a_{n} =^{2 n} C_{n}$
Since $^{2 n} C_{n}$ is greatest coefficient (because it is the coefficient of the middle term), hence obviously
$^{2 n} C_{n} >^{2 n} C_{n + 1}$
$a_{n} > a_{n + 1}$
Also
$a_{n + 3}$ $=^{2 n} C_{n + 3}$

$=^{2 n} C_{2 n - (n + 3)}$
$=^{2 n} C_{n - 3}$
$= a_{n - 3}$ .
Hence
$a_{n + 3} = a_{n - 3}$ .

Suggest Corrections

Similar questions

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . . . + a_{2 n} x^{2 n}

a_{0}, a_{1}, a_{2}, . . . . . . . . . . ., a_{2 n}

a_{n}

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}

a_{0}, a_{1}, a_{2}, . . ., a_{2 n}

a_{n} = \frac{1}{2 n + 1}

n \in N

(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . a_{2 n} x^{2 n}

a_{0} + a_{1} + a_{2} . . . . a_{n - 1}

If $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a^{2 n} x^{2 n}, t h e n a_{0} + a_{3} + a_{6} + . . . . . =$

(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}

a_{0} + a_{3} + a_{6} + . . . =

Join BYJU'S Learning Program