JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 90

Let $\{a_n\}_{n=0}^\infty$ be a sequence such that $a_0=a_1=0$ and $a_{n+2}=3a_{n+1}-2a_n+1$ , ∀ n ≥ 0.

Then $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to

483

528

575

624

Answer and worked solution

Official answer: B

Solve. $a_n=2^n-n-1$ , so $a_{n+1}-2a_n=n$ .

Factor. The expression is $(a_{25}-2a_{24})(a_{23}-2a_{22})=24\times22=528$ .

(B)

Topics:Special series sums

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 6