NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sequence and Series
118897
If sum of the series \(\sum_{n=0}^{\infty} r^n=S\), for \(|r|\lt 1\), then sum of the series \(\sum_{n=0}^{\infty} r^{2 n}\), is
1 \(\mathrm{S}^2\)
2 \(\frac{S^2}{2 S+1}\)
3 \(\frac{2 \mathrm{~S}}{\mathrm{~S}^2-1}\)
4 \(\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
Explanation:
D Given, \(\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{\mathrm{n}}\) \(\Rightarrow \quad 1+\mathrm{r}^1+\mathrm{r}^2+\mathrm{r}^3+\ldots \infty=\mathrm{S}\) Sum of infinite terms of G.P., \(\mathrm{S}=\frac{1}{1-\mathrm{r}} \Rightarrow \mathrm{S}(1-\mathrm{r})=1\) \(\mathrm{r}=1-\frac{1}{\mathrm{~S}} \Rightarrow \mathrm{r}=\left(\frac{\mathrm{S}-1}{\mathrm{~S}}\right)\) Let, \(\quad \mathrm{y}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}\) \(y=1+r^2+r^4+\ldots \infty\) \(y=\frac{1}{\left(1-r^2\right)} \quad \text { [Sum of infinite terms of G.P.] }\) Putting the value of \(r\) from equation (i), we get \(y=\frac{1}{1-\frac{(S-1)^2}{S^2}}\) \(y=\frac{S^2}{S^2-(S-1)^2}=\frac{S^2}{[S-(S-1)][S+S-1]}\) \(y=\frac{S^2}{(2 S-1)}\)So, \(\quad \sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}=\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
BCECE-2014
Sequence and Series
118898
Find the sum of the series \(\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots \ldots . .\)
1 \(\log \frac{\mathrm{e}}{2}\)
2 \(\log \frac{\mathrm{e}}{4}\)
3 \(\log \frac{2}{3}\)
4 \(\log \frac{2}{4}\)
Explanation:
A Let, sum of the given series be \(\mathrm{S}\). Then, \(\quad S=\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots\) \(\mathrm{S}=\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{4}-\frac{1}{5}\right]+\left[\frac{1}{6}-\frac{1}{7}\right]+\ldots . .\) \(\mathrm{S} =1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\) \(= 1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\) \(= 1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\ldots\right)\) \(= 1-\log 2\) \(= \log \mathrm{e}-\log 2=\log \frac{\mathrm{e}}{2}\)
JCECE-2012
Sequence and Series
118899
The sum \(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) is equal to
1 \(\frac{7}{87}\)
2 \(\frac{7}{29}\)
3 \(\frac{14}{87}\)
4 \(\frac{21}{29}\)
Explanation:
B \(\sum_{\mathrm{n}=1}^{21} \frac{3}{(4 \mathrm{n}-1)(4 \mathrm{n}+3)}\) Multiplication of \(\frac{1}{4}\) in equation (i) \(\quad=\frac{3}{4} \sum_{\mathrm{n}=1}^{21}\left(\frac{1}{(4 \mathrm{n}-1)}-\frac{1}{(4 \mathrm{n}+3)}\right)\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right) \cdots\left(\frac{1}{83}-\frac{1}{87}\right)\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7} \ldots . . \frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{29-1}{87}\right]\) \(=\frac{3}{4}\left[\frac{28}{87}\right]\) \(=\frac{7}{29}\)
JEE Main-25.07.2022
Sequence and Series
118900
If the sum of first \(n\) natural numbers is \(\mathbf{1 / 5}\) times the sum of their squares, then \(n=\)
118897
If sum of the series \(\sum_{n=0}^{\infty} r^n=S\), for \(|r|\lt 1\), then sum of the series \(\sum_{n=0}^{\infty} r^{2 n}\), is
1 \(\mathrm{S}^2\)
2 \(\frac{S^2}{2 S+1}\)
3 \(\frac{2 \mathrm{~S}}{\mathrm{~S}^2-1}\)
4 \(\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
Explanation:
D Given, \(\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{\mathrm{n}}\) \(\Rightarrow \quad 1+\mathrm{r}^1+\mathrm{r}^2+\mathrm{r}^3+\ldots \infty=\mathrm{S}\) Sum of infinite terms of G.P., \(\mathrm{S}=\frac{1}{1-\mathrm{r}} \Rightarrow \mathrm{S}(1-\mathrm{r})=1\) \(\mathrm{r}=1-\frac{1}{\mathrm{~S}} \Rightarrow \mathrm{r}=\left(\frac{\mathrm{S}-1}{\mathrm{~S}}\right)\) Let, \(\quad \mathrm{y}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}\) \(y=1+r^2+r^4+\ldots \infty\) \(y=\frac{1}{\left(1-r^2\right)} \quad \text { [Sum of infinite terms of G.P.] }\) Putting the value of \(r\) from equation (i), we get \(y=\frac{1}{1-\frac{(S-1)^2}{S^2}}\) \(y=\frac{S^2}{S^2-(S-1)^2}=\frac{S^2}{[S-(S-1)][S+S-1]}\) \(y=\frac{S^2}{(2 S-1)}\)So, \(\quad \sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}=\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
BCECE-2014
Sequence and Series
118898
Find the sum of the series \(\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots \ldots . .\)
1 \(\log \frac{\mathrm{e}}{2}\)
2 \(\log \frac{\mathrm{e}}{4}\)
3 \(\log \frac{2}{3}\)
4 \(\log \frac{2}{4}\)
Explanation:
A Let, sum of the given series be \(\mathrm{S}\). Then, \(\quad S=\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots\) \(\mathrm{S}=\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{4}-\frac{1}{5}\right]+\left[\frac{1}{6}-\frac{1}{7}\right]+\ldots . .\) \(\mathrm{S} =1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\) \(= 1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\) \(= 1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\ldots\right)\) \(= 1-\log 2\) \(= \log \mathrm{e}-\log 2=\log \frac{\mathrm{e}}{2}\)
JCECE-2012
Sequence and Series
118899
The sum \(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) is equal to
1 \(\frac{7}{87}\)
2 \(\frac{7}{29}\)
3 \(\frac{14}{87}\)
4 \(\frac{21}{29}\)
Explanation:
B \(\sum_{\mathrm{n}=1}^{21} \frac{3}{(4 \mathrm{n}-1)(4 \mathrm{n}+3)}\) Multiplication of \(\frac{1}{4}\) in equation (i) \(\quad=\frac{3}{4} \sum_{\mathrm{n}=1}^{21}\left(\frac{1}{(4 \mathrm{n}-1)}-\frac{1}{(4 \mathrm{n}+3)}\right)\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right) \cdots\left(\frac{1}{83}-\frac{1}{87}\right)\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7} \ldots . . \frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{29-1}{87}\right]\) \(=\frac{3}{4}\left[\frac{28}{87}\right]\) \(=\frac{7}{29}\)
JEE Main-25.07.2022
Sequence and Series
118900
If the sum of first \(n\) natural numbers is \(\mathbf{1 / 5}\) times the sum of their squares, then \(n=\)
118897
If sum of the series \(\sum_{n=0}^{\infty} r^n=S\), for \(|r|\lt 1\), then sum of the series \(\sum_{n=0}^{\infty} r^{2 n}\), is
1 \(\mathrm{S}^2\)
2 \(\frac{S^2}{2 S+1}\)
3 \(\frac{2 \mathrm{~S}}{\mathrm{~S}^2-1}\)
4 \(\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
Explanation:
D Given, \(\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{\mathrm{n}}\) \(\Rightarrow \quad 1+\mathrm{r}^1+\mathrm{r}^2+\mathrm{r}^3+\ldots \infty=\mathrm{S}\) Sum of infinite terms of G.P., \(\mathrm{S}=\frac{1}{1-\mathrm{r}} \Rightarrow \mathrm{S}(1-\mathrm{r})=1\) \(\mathrm{r}=1-\frac{1}{\mathrm{~S}} \Rightarrow \mathrm{r}=\left(\frac{\mathrm{S}-1}{\mathrm{~S}}\right)\) Let, \(\quad \mathrm{y}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}\) \(y=1+r^2+r^4+\ldots \infty\) \(y=\frac{1}{\left(1-r^2\right)} \quad \text { [Sum of infinite terms of G.P.] }\) Putting the value of \(r\) from equation (i), we get \(y=\frac{1}{1-\frac{(S-1)^2}{S^2}}\) \(y=\frac{S^2}{S^2-(S-1)^2}=\frac{S^2}{[S-(S-1)][S+S-1]}\) \(y=\frac{S^2}{(2 S-1)}\)So, \(\quad \sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}=\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
BCECE-2014
Sequence and Series
118898
Find the sum of the series \(\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots \ldots . .\)
1 \(\log \frac{\mathrm{e}}{2}\)
2 \(\log \frac{\mathrm{e}}{4}\)
3 \(\log \frac{2}{3}\)
4 \(\log \frac{2}{4}\)
Explanation:
A Let, sum of the given series be \(\mathrm{S}\). Then, \(\quad S=\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots\) \(\mathrm{S}=\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{4}-\frac{1}{5}\right]+\left[\frac{1}{6}-\frac{1}{7}\right]+\ldots . .\) \(\mathrm{S} =1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\) \(= 1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\) \(= 1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\ldots\right)\) \(= 1-\log 2\) \(= \log \mathrm{e}-\log 2=\log \frac{\mathrm{e}}{2}\)
JCECE-2012
Sequence and Series
118899
The sum \(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) is equal to
1 \(\frac{7}{87}\)
2 \(\frac{7}{29}\)
3 \(\frac{14}{87}\)
4 \(\frac{21}{29}\)
Explanation:
B \(\sum_{\mathrm{n}=1}^{21} \frac{3}{(4 \mathrm{n}-1)(4 \mathrm{n}+3)}\) Multiplication of \(\frac{1}{4}\) in equation (i) \(\quad=\frac{3}{4} \sum_{\mathrm{n}=1}^{21}\left(\frac{1}{(4 \mathrm{n}-1)}-\frac{1}{(4 \mathrm{n}+3)}\right)\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right) \cdots\left(\frac{1}{83}-\frac{1}{87}\right)\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7} \ldots . . \frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{29-1}{87}\right]\) \(=\frac{3}{4}\left[\frac{28}{87}\right]\) \(=\frac{7}{29}\)
JEE Main-25.07.2022
Sequence and Series
118900
If the sum of first \(n\) natural numbers is \(\mathbf{1 / 5}\) times the sum of their squares, then \(n=\)
118897
If sum of the series \(\sum_{n=0}^{\infty} r^n=S\), for \(|r|\lt 1\), then sum of the series \(\sum_{n=0}^{\infty} r^{2 n}\), is
1 \(\mathrm{S}^2\)
2 \(\frac{S^2}{2 S+1}\)
3 \(\frac{2 \mathrm{~S}}{\mathrm{~S}^2-1}\)
4 \(\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
Explanation:
D Given, \(\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{\mathrm{n}}\) \(\Rightarrow \quad 1+\mathrm{r}^1+\mathrm{r}^2+\mathrm{r}^3+\ldots \infty=\mathrm{S}\) Sum of infinite terms of G.P., \(\mathrm{S}=\frac{1}{1-\mathrm{r}} \Rightarrow \mathrm{S}(1-\mathrm{r})=1\) \(\mathrm{r}=1-\frac{1}{\mathrm{~S}} \Rightarrow \mathrm{r}=\left(\frac{\mathrm{S}-1}{\mathrm{~S}}\right)\) Let, \(\quad \mathrm{y}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}\) \(y=1+r^2+r^4+\ldots \infty\) \(y=\frac{1}{\left(1-r^2\right)} \quad \text { [Sum of infinite terms of G.P.] }\) Putting the value of \(r\) from equation (i), we get \(y=\frac{1}{1-\frac{(S-1)^2}{S^2}}\) \(y=\frac{S^2}{S^2-(S-1)^2}=\frac{S^2}{[S-(S-1)][S+S-1]}\) \(y=\frac{S^2}{(2 S-1)}\)So, \(\quad \sum_{\mathrm{n}=0}^{\infty} \mathrm{r}^{2 \mathrm{n}}=\frac{\mathrm{S}^2}{2 \mathrm{~S}-1}\)
BCECE-2014
Sequence and Series
118898
Find the sum of the series \(\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots \ldots . .\)
1 \(\log \frac{\mathrm{e}}{2}\)
2 \(\log \frac{\mathrm{e}}{4}\)
3 \(\log \frac{2}{3}\)
4 \(\log \frac{2}{4}\)
Explanation:
A Let, sum of the given series be \(\mathrm{S}\). Then, \(\quad S=\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots\) \(\mathrm{S}=\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{4}-\frac{1}{5}\right]+\left[\frac{1}{6}-\frac{1}{7}\right]+\ldots . .\) \(\mathrm{S} =1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\) \(= 1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\) \(= 1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\ldots\right)\) \(= 1-\log 2\) \(= \log \mathrm{e}-\log 2=\log \frac{\mathrm{e}}{2}\)
JCECE-2012
Sequence and Series
118899
The sum \(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) is equal to
1 \(\frac{7}{87}\)
2 \(\frac{7}{29}\)
3 \(\frac{14}{87}\)
4 \(\frac{21}{29}\)
Explanation:
B \(\sum_{\mathrm{n}=1}^{21} \frac{3}{(4 \mathrm{n}-1)(4 \mathrm{n}+3)}\) Multiplication of \(\frac{1}{4}\) in equation (i) \(\quad=\frac{3}{4} \sum_{\mathrm{n}=1}^{21}\left(\frac{1}{(4 \mathrm{n}-1)}-\frac{1}{(4 \mathrm{n}+3)}\right)\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right) \cdots\left(\frac{1}{83}-\frac{1}{87}\right)\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7} \ldots . . \frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]\) \(=\frac{3}{4}\left[\frac{29-1}{87}\right]\) \(=\frac{3}{4}\left[\frac{28}{87}\right]\) \(=\frac{7}{29}\)
JEE Main-25.07.2022
Sequence and Series
118900
If the sum of first \(n\) natural numbers is \(\mathbf{1 / 5}\) times the sum of their squares, then \(n=\)
