1
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $f(n)=A(-2)^n+B(-3)^n \forall A, B \in \mathbf{R}$ and $n \in \mathbf{N}-\{1,2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

A

0

B

5

C

7

D

11

2
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots \ldots=\frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in \mathbf{R}$, then $(a-b)^2=$

A

0

B

64

C

36

D

125

3
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $f(1)=3$, and $f(n+1)-f(n)=3\left(4^n-1\right)$, then $\forall n \in \mathbf{N}$, $f(n)=$

A

$4^n-1$

B

$4^n-5 n+4$

C

$4^n-3 n+2$

D

$4^n+4 n-5$

4
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty$, then, when $n$ is even $S_n=$

A

$\frac{n(n+1)}{2}$

B

$\frac{n^2(n+1)}{2}$

C

$\frac{n(n+1)^2}{2}$

D

$\frac{n^2(n+2)}{2}$

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