Sums of special series — NEET & JEE Questions

Question 1 · difficulty L2 · understanding

$\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)$ is equal to:

• A. 65
• B. 130
• C. 220
• D. 440

Question 2 · difficulty L2 · understanding

The sum $$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}}$$ is equal to

• A. $$\frac{7}{87}$$
• B. $$\frac{7}{29}$$
• C. $$\frac{14}{87}$$
• D. $$\frac{21}{29}$$

Question 3 · difficulty L2 · understanding

The value of $$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}}$$ is equal to:

• A. $${{20} \over {11}}$$
• B. $${{11} \over {6}}$$
• C. $${{241} \over {132}}$$
• D. $${{21} \over {11}}$$

Question 4 · difficulty L3 · understanding

The sum $1+\frac{1}{2}\left(1^2+2^2\right)+\frac{1}{3}\left(1^2+2^2+3^2\right)+\ldots$ upto 10 terms is equal to :

• A. 130
• B. 155
• C. $\frac{315}{2}$
• D. $\frac{325}{2}$

Question 5 · difficulty L3 · understanding

The value of $1^3-2^3+3^3-\ldots+15^3$ is:

• A. 1706
• B. 1856
• C. 1982
• D. 2403

Question 6 · difficulty L3 · understanding

If the sum of the first 10 terms of the series $\frac{1}{1+1^4 \times 4}+\frac{2}{1+2^4 \times 4}+\frac{3}{1+3^4 \times 4}+\frac{4}{1+4^4 \times 4}+\ldots \ldots$. is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to :

• A. 256
• B. 264
• C. 276
• D. 284

Question 7 · difficulty L3 · understanding

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots$ up to 8 terms, is :

• A. 70
• B. 71
• C. 72
• D. 73

Question 8 · difficulty L3 · understanding

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3}$ is equal to :

• A. $2^{26}$
• B. $3^{25}$
• C. $3^{26}$
• D. $2^{25}$

Question 9 · difficulty L3 · understanding

The value of $\sum\limits_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right)$ is

• A. e/2
• B. $\sqrt{e}$
• C. 2/e
• D. 1/e

Question 10 · difficulty L3 · understanding

$\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots$ upto infinite terms, is equal to

• A. $\frac{7}{4}$
• B. $\frac{4}{3}$
• C. $\frac{6}{5}$
• D. $\frac{5}{2}$