[JEE Main 2006] if 1 1 ax 1 bx a 0 a 1x a 2x 2 then
If $\dfrac1{(1-ax)(1-bx)}=a_0+a_1x+a_2x^2+\cdots$, then $a_n$ is
(a) $\dfrac{b^n-a^n}{b-a}$
(b) $\dfrac{a^n-b^n}{b-a}$
(c) $\dfrac{a^{n+1}-b^{n+1}}{b-a}$
(d) $\dfrac{b^{n+1}-a^{n+1}}{b-a}$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (d) $\dfrac{b^{n+1}-a^{n+1}}{b-a}$
Partial fractions give $a_n=\frac{a^{n+1}-b^{n+1}}{a-b}=\frac{b^{n+1}-a^{n+1}}{b-a}$.
JEE Main 2006 · Binomial Theorem — verified solution by the Vidaara Team.
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.