JEE Main 2025 Jan 28 Shift 1, Mathematics Q9: Summation of Series

JEE Main2025Jan 28, Shift 1Mathematics

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, \dots$ Then $k = 1 \sum 100 a_{k}$ is equal to :

A
$3 a_{99} - 100$
B
$3 a_{100} - 100$
C
$3 a_{100} + 100$
D
$3 a_{99} + 100$

Solution

The recurrence $2 a_{n + 2} - 5 a_{n + 1} + 3 a_{n} = 0$ has characteristic equation $2 x^{2} - 5 x + 3 = 0$ , with roots $x = 1$ and $x = \frac{3}{2}$ .

General solution: $a_{n} = A (1)^{n} + B (\frac{3}{2})^{n}$ .

From $a_{0} = 0$ : $A + B = 0$ . From $a_{1} = \frac{1}{2}$ : $A + \frac{3}{2} B = \frac{1}{2}$ . Solving: $B = 1$ , $A = - 1$ .

$∴ a_{n} = (\frac{3}{2})^{n} - 1$ .

Sum: $k = 1 \sum 100 a_{k} = k = 1 \sum 100 (\frac{3}{2})^{k} - 100$ .

The geometric part: $\sum_{k = 1}^{100} (\frac{3}{2})^{k} = \frac{( 3/2 ) [ ( 3/2 ) ^{100} - 1 ]}{( 3/2 ) - 1} = 3 [(\frac{3}{2})^{100} - 1]$ .

Since $(\frac{3}{2})^{100} = a_{100} + 1$ :

$\sum_{k = 1}^{100} (\frac{3}{2})^{k} = 3 a_{100}$ .

$∴ \sum_{k = 1}^{100} a_{k} = 3 a_{100} - 100$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q9: Summation of Series

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