MCQ

MCQ 11 Mark

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

Correct option: C.

$\text{n}-1-\frac{1}{2^{\text{n}}}$

We have,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1(2^{\text{r}}-1)}{2\text{r}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\Big(1-\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\sum\limits^{\text{n}}_{\text{r}=1}\Big(\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Bigg[\frac{\big(\frac{1}{2}\big)\big\{1-\big(\frac{1}{2}\big)^\text{n}\big\}}{1-\frac{1}{2}}\Bigg]$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Big[1-\Big(\frac{1}{2}\Big)^\text{n}\Big]$
$\Rightarrow\text{S}_\text{n}=\text{n}-1+\frac{1}{2^\text{n}}$

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MCQ 21 Mark

The sum of the series $\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}$ 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

Correct option: C.

$\frac{\text{n}(\text{n}+1)}{4}$

Let $\text{S}_\text{n}=\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log4}{\log4}+\frac{\log8}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log2^2}{\log4}+\frac{\log2^3}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{2\log2}{\log4}+\frac{3\log2}{\log4}+\ ...\ +\frac{\text{n}\log2}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\log4^{\frac{1}{2}}}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\frac{1}{2}\log4}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{1}{2}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}$

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MCQ 31 Mark

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

Correct option: C.

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

Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\sqrt{2\times\text{n}^2}=\text{n}\sqrt{2}$
Now, let $S_n$ be the sum of n terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sqrt{2}\sum\limits^{\text{n}}_{\text{k}=1}(\text{k})$
$\Rightarrow\text{S}_\text{n}=\sqrt{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$

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MCQ 41 Mark

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

Correct option: B.

$\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$

We have,
$\sum\limits^{\text{n}}_{\text{r}=1}\Big\{(2\text{r}-1)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}\Big\{2\text{ra}-\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}2\text{ar}-\sum\limits^{\text{n}}_{\text{r}=1}\text{a}+\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{\text{b}^{\text{r}}}$
$=\text{an}(\text{n}+1)-\text{an}+\frac{(1-\text{b}^\text{n})}{(1-\text{b})\text{b}^\text{n}}$
$=\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$

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MCQ 51 Mark

The sum of $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ ...\text{ is}$

✓
$121\big(\sqrt{6}+\sqrt{2}\big)$
B
$243\big(\sqrt{3}+1\big)$
C
$\frac{121}{\sqrt{3}-1}$
D
$242\big(\sqrt{3}-1\big)$

Answer

Correct option: A.

$121\big(\sqrt{6}+\sqrt{2}\big)$

Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{S}_{10}=\sqrt{2}\sum\limits^{10}_{\text{k}=1}\Big(\sqrt{3^{(\text{k}-1)}}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\Big(1+\sqrt{3}+\sqrt{3^2}+\ ...\ +\sqrt{3^9}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\bigg(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\bigg)$
$\Rightarrow\text{S}_{10\text{n}}=\sqrt{2}\Big(\frac{3^5-1}{\sqrt{3}-1}\Big)\Big(\frac{\sqrt{3}+1}{\sqrt{3}+1}\Big)$
$\Rightarrow\text{S}_{10}=\frac{\sqrt{2}}{2}\big(3^5-1\big)\big(\sqrt{3}+1\big)$
$\Rightarrow\text{S}_{10}=\frac{1}{2}(242)\Big(\sqrt{6}+\sqrt{2}\Big)$
$\Rightarrow\text{S}_{10}=121\big(\sqrt{6}+\sqrt{2}\big)$

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MCQ 61 Mark

The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ ....$ to $n$ terms is:

A
$\text{n}-\frac{1}{2}(3^{-\text{n}}-1)$
✓
$\text{n}-\frac{1}{2}(1-3^{-\text{n}})$
C
$\text{n}+\frac{1}{2}(3^\text{n}-1)$
D
$\text{n}-\frac{1}{1}(3^\text{n}-1)$

Answer

Correct option: B.

$\text{n}-\frac{1}{2}(1-3^{-\text{n}})$

Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{3^\text{n}-1}{3^\text{n}}=1-\frac{1}{3^\text{n}}$
Now,
Let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum^\limits\text{n}_{\text{k}=1}\text{T}_\text{k}$
$=\sum^\limits\text{n}_{\text{k}=1}\Big[1-\frac{1}{3^\text{k}}\Big]$
$=\sum^\limits\text{n}_{\text{k}=1}1-\sum^\limits\text{n}_{\text{k}=1}\frac{1}{3^\text{k}}$
$=\text{n}-\Big[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ ....\ +\frac{1}{3^\text{n}}\Big]$
$=\text{n}-\frac{1}{3}\Bigg[\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac{1}{3}}\Bigg]$
$=\text{n}-\frac{1}{2}\Big[1-\Big(\frac{1}{3}\Big)^\text{n}\Big]$
$=\text{n}-\frac{1}{3}\big[1-3^{-\text{n}}\big]$

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MCQ 71 Mark

If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ ....$ to n terms is $S,$ then $S$ is equal to:

✓
$\frac{\text{n}(\text{n}+3)}{4}$
B
$\frac{\text{n}(\text{n}+2)}{4}$
C
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
D
$\text{n}^2$

Answer

Correct option: A.

$\frac{\text{n}(\text{n}+3)}{4}$

Let $T_n$ be the nth term of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}}{2}+\frac{1}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{\text{k}}{2}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+1}{2}+1\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+3}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+3)}{4}$

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MCQ 81 Mark

The sum of the series $1^2 + 3^2 + 5^2 + ...$ to $n$ terms is:

A
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{2}$
✓
$\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
C
$\frac{(\text{n}-1)^2(2\text{n}+1)}{6}$
D
$\frac{(2\text{n}+1)^3}{3}$

Answer

Correct option: B.

$\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$

Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=(2\text{n}-1)^2$
$=4\text{n}^2+1-4\text{n}$
Now, let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_\text{k=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_\text{k=1}\text{k}^2+\sum\limits^{\text{n}}_\text{k=1}1-4\sum\limits^{\text{n}}_\text{k=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}(\text{n}+1)(2\text{n}+1)}{3}+\text{n}-2\text{n}(\text{n}+1)$
$\Rightarrow\text{S}_\text{n}=\text{n}\Big[\frac{2(\text{n}+1)(2\text{n}+1)}{3}+1-2(\text{n}+1)\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(2\text{n}+2)(2\text{n}+1)+3-6(\text{n}+1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(4\text{n}^2-1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$

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MCQ 91 Mark

If $\sum\text{n}=210,$ then $\sum\text{n}^2=$

✓
$2870$
B
$2160$
C
$2970$
D
none of these.

Answer

Correct option: A.

$2870$

Given,
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$
$(\because\ \text{n}>0)$
Now,
$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}\times\frac{(2\text{n}+1)}{3}$
$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$
$\Rightarrow(70)\times(41)$
$\Rightarrow2870$

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MCQ 101 Mark

The sum to n terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ .... $ is :

A
$\sqrt{2\text{n}+1}$
B
$\frac{1}{2}\sqrt{2\text{n}+1}$
C
$\sqrt{2\text{n}+1}-1$
✓
$\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$

Answer

Correct option: D.

$\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$

Let $T_n$ be the nth term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{1}{\sqrt{2\text{n}-1}+\sqrt{2\text{n}+1}}$
$=\frac{\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}}{2}$
Now,
Let $S_n$ be the sum $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$=\sum\limits^{\text{n}}_{\text{k}=1}\bigg(\frac{\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}}{2}\bigg)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\big(\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}\big)$
$=\frac{1}{2}\Big[\big(\sqrt{3}-\sqrt{1}\big)+\big(\sqrt{5}-\sqrt{3}\big)+\big(\sqrt{7}-\sqrt{5}\big)+\ ...\ +\big(\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}\big)\Big]$
$=\frac{1}{2}\Big\{(-1)+\sqrt{2\text{n}+1}\Big\}$
$=\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$

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