1
VITEEE 2024
MCQ (Single Correct Answer)
+1
-0

Sum of first ' $n$ ' terms of a series $a_1+a_2+\ldots+a_n$ is given by $S_n=\frac{n\left(n^2-1\right)(n+2)}{4}$, then the value of $\lim _\limits{n \rightarrow \infty} \sum_\limits{r=2}^n \frac{1}{a_r}$ is

A
4
B
2
C
$\frac{1}{4}$
D
$\frac{1}{2}$
2
VITEEE 2024
MCQ (Single Correct Answer)
+1
-0

The sum of $1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots \infty$ is

A
$\sqrt{3}$
B
$\frac{1}{\sqrt{2}}$
C
$\sqrt{2}$
D
$2^{\frac{3}{2}}$
3
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

Let $$a_n$$ be a sequence of numbers which is defined by relation $$a_1=2, \frac{a_n}{a_{n+1}}=3^{-n}$$, then $$\log _2\left(a_{50}\right)$$ is equal to (take $$\log _2 3=1.6$$ )

A
1960
B
1275
C
1961
D
1276
4
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

The value of $$\frac{1}{2}\left(\frac{1}{5}\right)^2+\frac{2}{3}\left(\frac{1}{5}\right)^3+\frac{3}{4}\left(\frac{1}{5}\right)^4+\ldots . . \infty$$ is

A
$$\frac{1}{4}+\log _e\left(\frac{4}{5}\right)$$
B
$$\frac{1}{20}+\log _e\left(\frac{4}{5}\right)$$
C
$$\frac{1}{20}-\log _e\left(\frac{4}{5}\right)$$
D
$$\frac{1}{20}+\log _e\left(\frac{6}{5}\right)$$
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