If $(1 + x + x^{2})^{25} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{50} x^{50}$ then $a_{0} + a_{2} + a_{4} + . . . . + a_{50}$ is

even

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

odd & of the form $3 n$

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

odd & of the form $3 n - 1$

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

odd & of the form $3 n + 1$

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

Open in App

Solution

The correct option is A
even

$(1 + x + x^{2})^{25} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{50} x^{50}$

Putting $x = 1$

$3^{25} = a_{0} + a_{1} + a_{2} + . . . + a_{50}$ ... (1)

Putting $x = - 1$

$1 = a_{0} - a_{1} + a_{2} - a_{3} + . . . + a_{50}$ ... (2)

Adding equations (1) and (2)

$3^{25} + 1 = 2 (a_{0} + a_{2} + a_{4} + . . . + a_{50})$

$\Rightarrow a_{0} + a_{2} + a_{4} + . . . . + a_{50} = \frac{1}{2} (3^{25} + 1)$

$= \frac{1}{2} [{(4 - 1)}^{25} + 1]$

$= \frac{1}{2} [(^{25} C_{0} 4^{25} -^{25} C_{1} 4^{24} + . . . +^{25} C_{24} 4^{1} -^{25} C_{25} 4^{0}) + 1]$

$= \frac{4}{2} [(^{25} C_{0} 4^{24} -^{25} C_{1} 4^{23} + . . . +^{25} C_{24} 4^{0})]$

$= 2 n$ which is always even. $[n ϵ N$ ]

Suggest Corrections

Similar questions

{(1 + x + x^{2})}^{25} = a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{50} \cdot x^{50}

a_{0} + a_{2} + a_{4} + \cdot \cdot \cdot + a_{50}

If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . a_{40} x^{40}$ , then $a_{0} + a_{2} + a_{4} . . . . a_{38}$ =

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

a_{0} + a_{2} + a_{4} + . . . . + a_{2 n}

(x + 10)^{50} + (x - 10)^{50} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{50} x^{50}

x \in R

\frac{a_{2}}{a_{0}}

(x + 10)^{50} + (x - 10)^{50} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{50} x^{50}

x \in R

\frac{a_{2}}{a_{0}}

Join BYJU'S Learning Program