Question types

Some Special Series question types

40 questions across 3 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

40
Questions
3
Question groups
5
Question types
Sample Questions

Some Special Series questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

If $\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}},$ then $S_n$ is equal to:
  • A
    $2^{\text{n}}-\text{n}-1$
  • B
    $1-\frac{1}{2^{\text{n}}}$
  • $\text{n}-1-\frac{1}{2^{\text{n}}}$
  • D
    ${2^{\text{n}}}-1$

Answer: C.

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The sum of the series $\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}\text{ is}:$
  • A
    $\frac{\text{n}(\text{n}+1)}{2}$
  • B
    $\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{12}$
  • $\frac{\text{n}(\text{n}+1)}{4}$
  • D
    none of these.

Answer: C.

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Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\text{ is}$
  • A
    $\frac{\text{n}(\text{n}+1)}{2}$
  • B
    $2\text{n}(\text{n}+1)$
  • $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
  • D
    $1$

Answer: C.

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The value of $\sum\limits^{\text{n}}_{\text{r}=1}\Big\{\big(2\text{r}-1\big)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$ is equal to:
  • A
    $\text{an}^2+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}-1}(\text{b}-1)}$
  • $\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
  • C
    $\text{an}^3+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}}(\text{b}-1)}$
  • D
    none of these.

Answer: B.

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The sum of $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+ ....$ is
  • $121\big(\sqrt{6}+\sqrt{2}\big)$
  • B
    $243\big(\sqrt{3}+1\big)$
  • C
    $\frac{121}{\sqrt{3}-1}$
  • D
    $242\big(\sqrt{3}-1\big)$

Answer: A.

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Let $S_n$ denote the sum of the cubes of first n natural numbers and $s_n$ denote the sum of first $n$ natural numbers. Then, write the value of $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$
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