If 1+x+x2n=a0+a1x+a2x2+…+a2nx2n, then a0+a2+a4+…+a2n=

Given 1+x+x2n=a0+a1x+a2x2+…+a2nx2n→(1) Put x=1 in (1). Then a0+a1+a2+a3+a4+…+a2=3n→(2) Put x=−1 in (1). Then a0−a1+a2−a3+a4+…+a2n=1→ (3) (2)+(3)⇒2a0+a2+a4+…+a2n=3n+1⇒a0+a2+a4+…+a2n=3n+12

If 1+x+x2n=a0+a1x+a2x2+…+a2nx2n, then a0+a2+a4+…+a2n=

Given 1+x+x2n=a0+a1x+a2x2+…+a2nx2n→(1) Put x=1 in (1). Then a0+a1+a2+a3+a4+…+a2=3n→(2) Put x=−1 in (1). Then a0−a1+a2−a3+a4+…+a2n=1→ (3) (2)+(3)⇒2a0+a2+a4+…+a2n=3n+1⇒a0+a2+a4+…+a2n=3n+12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}, then a_{0} + a_{2} + a_{4} + \dots + a_{2 n} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3^{n}$

$3^{n} + 1$

$\frac{3^{n} - 1}{2}$

$\frac{3^{n} + 1}{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

$Put x = 1 in (1). Then a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + \dots + a_{2} = 3^{n} \to (2)$

$Put x = - 1 in (1). Then a_{0} - a_{1} + a_{2} - a_{3} + a_{4} + \dots + a_{2 n} = 1 \to (3)$

$(2) + (3) \Rightarrow 2 (a_{0} + a_{2} + a_{4} + \dots + a_{2 n}) = 3^{n} + 1 \Rightarrow a_{0} + a_{2} + a_{4} + \dots + a_{2 n} = \frac{3^{n} + 1}{2}$