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

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

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 $3 a_{99} + 100$

Ans.
(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

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

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

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

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

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

$= 3 a_{100} - 100$ .

Want Us to Email you About Special Offers & Updates?