Let be a sequence such that and . Then is equal to . [2025]

Question

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

Smart Achievers · Accepted Answer

(2)

We have, $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$

Let $2 x^{2} = 5 x - 3$

$\Rightarrow 2 x^{2} - 5 x + 3 = 0 \Rightarrow (2 x - 3) (x - 1) = 0$

$\Rightarrow x = 1, \frac{3}{2} ∴ a_{n} = A {(1)}^{n} + B {(\frac{3}{2})}^{n}$

Put n = 0; 0 = A + B

$n = 1; \frac{1}{2} = A + \frac{3}{2} B \Rightarrow B = 1, A = - 1$

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

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

$= - 100 + \frac{(\frac{3}{2}) ({(\frac{3}{2})}^{100} - 1)}{(\frac{3}{2} - 1)}$

$\Rightarrow - 100 + [3 ({(\frac{3}{2})}^{100} - 1)]$

$= 3 a_{100} - 100$ .

Solution

Q.
Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

1 $3 a_{99} - 100$

2 $3 a_{100} - 100$

3 $3 a_{100} + 100$

4 $3 a_{99} + 100$

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Solution

Q. Let <an> be a sequence such that a0=0,a1=12 and 2an+2=5an+1–3an, n=0,1,2,3,.... Then ∑k=1100ak is equal to . [2025]

1 3a99–100

2 3a100–100

3 3a100+100