MATHS M.C.Q (1 Marks)

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

Solution:

We have,

$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}}$

$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1(2^{\text{r}}-1)}{2\text{r}}$

$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\Big(1-\frac{1}{2^{\text{r}}}\Big)$

$\Rightarrow\text{S}_\text{n}=\text{n}-\sum\limits^{\text{n}}_{\text{r}=1}\Big(\frac{1}{2^{\text{r}}}\Big)$

$\Rightarrow\text{S}_\text{n}=\text{n}-\Bigg[\frac{\big(\frac{1}{2}\big)\big\{1-\big(\frac{1}{2}\big)^\text{n}\big\}}{1-\frac{1}{2}}\Bigg]$

$\Rightarrow\text{S}_\text{n}=\text{n}-\Big[1-\Big(\frac{1}{2}\Big)^\text{n}\Big]$

$\Rightarrow\text{S}_\text{n}=\text{n}-1+\frac{1}{2^\text{n}}$

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

Solution:

Let $\text{S}_\text{n}=\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log4}{\log4}+\frac{\log8}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log2^2}{\log4}+\frac{\log2^3}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{2\log2}{\log4}+\frac{3\log2}{\log4}+\ ...\ +\frac{\text{n}\log2}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\log4^{\frac{1}{2}}}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\frac{1}{2}\log4}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{1}{2}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}$

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

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

$\text{S}_\text{n}=\sqrt{2}\sum\limits^{\text{n}}_{\text{k}=1}(\text{k})$

$\Rightarrow\text{S}_\text{n}=\sqrt{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$

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

Solution:

We have,

$\sum\limits^{\text{n}}_{\text{r}=1}\Big\{(2\text{r}-1)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$

$=\sum\limits^{\text{n}}_{\text{r}=1}\Big\{2\text{ra}-\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$

$=\sum\limits^{\text{n}}_{\text{r}=1}2\text{ar}-\sum\limits^{\text{n}}_{\text{r}=1}\text{a}+\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{\text{b}^{\text{r}}}$

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

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

1. $121\big(\sqrt{6}+\sqrt{2}\big)$

Solution:

Let T_n be the n^th term of the given series.

Thus, we have

$\text{S}_{10}=\sqrt{2}\sum\limits^{10}_{\text{k}=1}\Big(\sqrt{3^{(\text{k}-1)}}\Big)$

$\Rightarrow\text{S}_{10}=\sqrt{2}\Big(1+\sqrt{3}+\sqrt{3^2}+\ ...\ +\sqrt{3^9}\Big)$

$\Rightarrow\text{S}_{10}=\sqrt{2}\bigg(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\bigg)$

$\Rightarrow\text{S}_{10\text{n}}=\sqrt{2}\Big(\frac{3^5-1}{\sqrt{3}-1}\Big)\Big(\frac{\sqrt{3}+1}{\sqrt{3}+1}\Big)$

$\Rightarrow\text{S}_{10}=\frac{\sqrt{2}}{2}\big(3^5-1\big)\big(\sqrt{3}+1\big)$

$\Rightarrow\text{S}_{10}=\frac{1}{2}(242)\Big(\sqrt{6}+\sqrt{2}\Big)$

$\Rightarrow\text{S}_{10}=121\big(\sqrt{6}+\sqrt{2}\big)$

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

Solution:

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

$\text{S}_\text{n}=\sum^\limits\text{n}_{\text{k}=1}\text{T}_\text{k}$

$=\sum^\limits\text{n}_{\text{k}=1}\Big[1-\frac{1}{3^\text{k}}\Big]$

$=\sum^\limits\text{n}_{\text{k}=1}1-\sum^\limits\text{n}_{\text{k}=1}\frac{1}{3^\text{k}}$

$=\text{n}-\Big[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ ....\ +\frac{1}{3^\text{n}}\Big]$

$=\text{n}-\frac{1}{3}\Bigg[\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac{1}{3}}\Bigg]$

$=\text{n}-\frac{1}{2}\Big[1-\Big(\frac{1}{3}\Big)^\text{n}\Big]$

$=\text{n}-\frac{1}{3}\big[1-3^{-\text{n}}\big]$

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

Solution:

Let T_n be the nth term of the given series.

Thus, we have

$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}}{2}+\frac{1}{2}\Big)$

$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{\text{k}}{2}+\frac{\text{n}}{2}$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+1}{2}+1\Big)$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+3}{2}\Big)$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+3)}{4}$

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

Solution:

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

$\text{S}_\text{n}=\sum\limits^{\text{n}}_\text{k=1}(4\text{k}^2+1-4\text{k})$

$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_\text{k=1}\text{k}^2+\sum\limits^{\text{n}}_\text{k=1}1-4\sum\limits^{\text{n}}_\text{k=1}\text{k}$

$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$

$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}(\text{n}+1)(2\text{n}+1)}{3}+\text{n}-2\text{n}(\text{n}+1)$

$\Rightarrow\text{S}_\text{n}=\text{n}\Big[\frac{2(\text{n}+1)(2\text{n}+1)}{3}+1-2(\text{n}+1)\Big]$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(2\text{n}+2)(2\text{n}+1)+3-6(\text{n}+1)\big]$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(4\text{n}^2-1)\big]$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$

1. 2870

Solution:

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,

$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$

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

$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$

$\Rightarrow(70)\times(41)$

$\Rightarrow2870$

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

Solution:

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

$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$

$=\sum\limits^{\text{n}}_{\text{k}=1}\bigg(\frac{\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}}{2}\bigg)$

$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\big(\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}\big)$

$=\frac{1}{2}\Big[\big(\sqrt{3}-\sqrt{1}\big)+\big(\sqrt{5}-\sqrt{3}\big)+\big(\sqrt{7}-\sqrt{5}\big)+\ ...\ +\big(\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}\big)\Big]$

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

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