JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
Let $\langle a_n\rangle$ be a sequence such that $a_0=0$, $a_1=\frac{1}{2}$ and $2a_{n+2}=5a_{n+1}-3a_n$, $n=0,1,2,3,\ldots$.
Then $\sum_{k=1}^{100}a_k$ is equal to
Sequence and Series · Logarithmic and Exponential Series, and Recursive Sequences · Geometric Progression
Solution
Formulas used. (1) For $2a_{n+2}=5a_{n+1}-3a_n$, try $a_n=x^n$: $2x^2-5x+3=0$(2) Distinct roots $x_1,x_2$ give $a_n=A\,x_1^n+B\,x_2^n$(3) Sum of a G.P.: $x+x^2+\cdots+x^N=x\,\frac{x^N-1}{x-1}$Step 1 - Roots. $2x^2-5x+3=(x-1)(2x-3)=0$, so $x=1$ or $x=\tfrac32$: $$a_n=A+B\left(\tfrac32\right)^n .$$ Step 2 - Constants. $a_0=A+B=0$ and $a_1=A+\tfrac32B=\tfrac12$ give $B=1,\ A=-1$: $$a_n=\left(\tfrac32\right)^n-1 .$$ Step 3 - The sum. $$\sum_{k=1}^{100}a_k=\sum_{k=1}^{100}\left(\tfrac32\right)^k-100=\frac32\cdot\frac{\left(\tfrac32\right)^{100}-1}{\tfrac12}-100=3\left[\left(\tfrac32\right)^{100}-1\right]-100 .$$ Step 4 - In terms of $a_{100}$. The bracket is exactly $a_{100}$: $$\sum_{k=1}^{100}a_k=3a_{100}-100 .$$ Answer: (D)