D Let, \(\mathrm{T}_{\mathrm{r}} =\mathrm{r}(\mathrm{r}+1)(\mathrm{r}+2)\) \(=\mathrm{r}\left(\mathrm{r}^2+3 \mathrm{r}+2\right)\) \(=\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\) Required sum of \(T_{\mathrm{r}}\) is \(\sigma=\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)\) As we know that \(\sum \mathrm{n}^3=\left[\frac{\mathrm{n}+(\mathrm{n}+1)}{2}\right]^2 \Rightarrow \sum \mathrm{n}^3=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) \(\sum \mathrm{n}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) \(\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)=\) \(\left(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+3 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\) \(=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+\frac{\mathrm{n}(2 \mathrm{n}+1)(\mathrm{n}+1)}{2}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{1}+2(2 \mathrm{n}+1)+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\mathrm{n}^2+\mathrm{n}+4 \mathrm{n}+2+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left(\mathrm{n}^2+5 \mathrm{n}+6\right)\) \(=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\)
AP EAMCET-2015
Sequence and Series
118902
Sum of the first 100 terms of the series \(1+3+7\) \(+\mathbf{1 5}+\mathbf{3 1}+\ldots .\). is
1 \(2^{100}-102\)
2 \(2^{101}-102\)
3 \(2^{102}-103\)
4 \(2^{102}-104\)
Explanation:
B As per give summation, the successive difference are \(2,4,8,16, \ldots\). which are in A.P. here, \(\mathrm{S}=1+3+7+15+31+\ldots \ldots+\mathrm{T}_{100}\) \(=\left(1^2-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\ldots \ldots+\left(2^{100}-1\right)\) \(=\left(2+2^2+2^3+\ldots .+2^{100}\right)-100\) \(=2\left[\frac{2^{100}-1}{2-1}\right]-100\) \(=2^{101}-102\)
J and K CET-2015
Sequence and Series
118903
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots . .(\mathrm{n}\) terms \()=\frac{\mathrm{kn}}{\mathrm{n}+1}\), then \(\mathrm{k}\) is equal to
1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
A Here, given series can be re-written as \(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots .+\frac{1}{2 n}-\frac{1}{2 n+2}\right]\) \(=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}}{\mathrm{n}+1}\right]\) On comparing given equation, we get. \(\frac{\mathrm{kn}}{\mathrm{n}+1}=\frac{1}{4} \frac{\mathrm{n}}{(\mathrm{n}+1)}\) \(\mathrm{k}=\frac{1}{4}\)
AP EAMCET-2012
Sequence and Series
118904
For all \(n \in \mathbb{N}\), if \(\mathbf{1}^2+2^2+3^2+\ldots \ldots .+n^2>x\), then \(\mathbf{x}=\)
1 \(\frac{\mathrm{n}^3}{3}\)
2 \(\frac{\mathrm{n}^3}{2}\)
3 \(\mathrm{n}^3\)
4 \(\frac{n^4}{4}\)
Explanation:
A Given that, \(1^2+2^2+3^2+\ldots . . n^2>x\) \(\sum_{n=1}^n n^2>x\) We know that, \(\sum n^2=\frac{n(n+1)(2 n+1)}{6}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\mathrm{x}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\frac{\mathrm{n}(\mathrm{n})(2 \mathrm{n}) .}{6}\) \(\mathrm{x}=\frac{\mathrm{n} \times \mathrm{n} \times 2 \mathrm{n}}{6}\) \(\mathrm{x}=\frac{\mathrm{n}^3}{3}\)
TS EAMCET-03.05.2019
Sequence and Series
118905
If \(f: R \rightarrow R\) is defined as \(f(x+y)=f(x)+f(y)\), \(\forall x, y \in R\) and \(f(1)=10\), then, \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{f}(\mathrm{r}))^2=\)
D Let, \(\mathrm{T}_{\mathrm{r}} =\mathrm{r}(\mathrm{r}+1)(\mathrm{r}+2)\) \(=\mathrm{r}\left(\mathrm{r}^2+3 \mathrm{r}+2\right)\) \(=\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\) Required sum of \(T_{\mathrm{r}}\) is \(\sigma=\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)\) As we know that \(\sum \mathrm{n}^3=\left[\frac{\mathrm{n}+(\mathrm{n}+1)}{2}\right]^2 \Rightarrow \sum \mathrm{n}^3=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) \(\sum \mathrm{n}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) \(\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)=\) \(\left(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+3 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\) \(=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+\frac{\mathrm{n}(2 \mathrm{n}+1)(\mathrm{n}+1)}{2}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{1}+2(2 \mathrm{n}+1)+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\mathrm{n}^2+\mathrm{n}+4 \mathrm{n}+2+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left(\mathrm{n}^2+5 \mathrm{n}+6\right)\) \(=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\)
AP EAMCET-2015
Sequence and Series
118902
Sum of the first 100 terms of the series \(1+3+7\) \(+\mathbf{1 5}+\mathbf{3 1}+\ldots .\). is
1 \(2^{100}-102\)
2 \(2^{101}-102\)
3 \(2^{102}-103\)
4 \(2^{102}-104\)
Explanation:
B As per give summation, the successive difference are \(2,4,8,16, \ldots\). which are in A.P. here, \(\mathrm{S}=1+3+7+15+31+\ldots \ldots+\mathrm{T}_{100}\) \(=\left(1^2-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\ldots \ldots+\left(2^{100}-1\right)\) \(=\left(2+2^2+2^3+\ldots .+2^{100}\right)-100\) \(=2\left[\frac{2^{100}-1}{2-1}\right]-100\) \(=2^{101}-102\)
J and K CET-2015
Sequence and Series
118903
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots . .(\mathrm{n}\) terms \()=\frac{\mathrm{kn}}{\mathrm{n}+1}\), then \(\mathrm{k}\) is equal to
1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
A Here, given series can be re-written as \(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots .+\frac{1}{2 n}-\frac{1}{2 n+2}\right]\) \(=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}}{\mathrm{n}+1}\right]\) On comparing given equation, we get. \(\frac{\mathrm{kn}}{\mathrm{n}+1}=\frac{1}{4} \frac{\mathrm{n}}{(\mathrm{n}+1)}\) \(\mathrm{k}=\frac{1}{4}\)
AP EAMCET-2012
Sequence and Series
118904
For all \(n \in \mathbb{N}\), if \(\mathbf{1}^2+2^2+3^2+\ldots \ldots .+n^2>x\), then \(\mathbf{x}=\)
1 \(\frac{\mathrm{n}^3}{3}\)
2 \(\frac{\mathrm{n}^3}{2}\)
3 \(\mathrm{n}^3\)
4 \(\frac{n^4}{4}\)
Explanation:
A Given that, \(1^2+2^2+3^2+\ldots . . n^2>x\) \(\sum_{n=1}^n n^2>x\) We know that, \(\sum n^2=\frac{n(n+1)(2 n+1)}{6}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\mathrm{x}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\frac{\mathrm{n}(\mathrm{n})(2 \mathrm{n}) .}{6}\) \(\mathrm{x}=\frac{\mathrm{n} \times \mathrm{n} \times 2 \mathrm{n}}{6}\) \(\mathrm{x}=\frac{\mathrm{n}^3}{3}\)
TS EAMCET-03.05.2019
Sequence and Series
118905
If \(f: R \rightarrow R\) is defined as \(f(x+y)=f(x)+f(y)\), \(\forall x, y \in R\) and \(f(1)=10\), then, \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{f}(\mathrm{r}))^2=\)
D Let, \(\mathrm{T}_{\mathrm{r}} =\mathrm{r}(\mathrm{r}+1)(\mathrm{r}+2)\) \(=\mathrm{r}\left(\mathrm{r}^2+3 \mathrm{r}+2\right)\) \(=\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\) Required sum of \(T_{\mathrm{r}}\) is \(\sigma=\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)\) As we know that \(\sum \mathrm{n}^3=\left[\frac{\mathrm{n}+(\mathrm{n}+1)}{2}\right]^2 \Rightarrow \sum \mathrm{n}^3=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) \(\sum \mathrm{n}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) \(\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)=\) \(\left(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+3 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\) \(=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+\frac{\mathrm{n}(2 \mathrm{n}+1)(\mathrm{n}+1)}{2}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{1}+2(2 \mathrm{n}+1)+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\mathrm{n}^2+\mathrm{n}+4 \mathrm{n}+2+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left(\mathrm{n}^2+5 \mathrm{n}+6\right)\) \(=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\)
AP EAMCET-2015
Sequence and Series
118902
Sum of the first 100 terms of the series \(1+3+7\) \(+\mathbf{1 5}+\mathbf{3 1}+\ldots .\). is
1 \(2^{100}-102\)
2 \(2^{101}-102\)
3 \(2^{102}-103\)
4 \(2^{102}-104\)
Explanation:
B As per give summation, the successive difference are \(2,4,8,16, \ldots\). which are in A.P. here, \(\mathrm{S}=1+3+7+15+31+\ldots \ldots+\mathrm{T}_{100}\) \(=\left(1^2-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\ldots \ldots+\left(2^{100}-1\right)\) \(=\left(2+2^2+2^3+\ldots .+2^{100}\right)-100\) \(=2\left[\frac{2^{100}-1}{2-1}\right]-100\) \(=2^{101}-102\)
J and K CET-2015
Sequence and Series
118903
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots . .(\mathrm{n}\) terms \()=\frac{\mathrm{kn}}{\mathrm{n}+1}\), then \(\mathrm{k}\) is equal to
1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
A Here, given series can be re-written as \(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots .+\frac{1}{2 n}-\frac{1}{2 n+2}\right]\) \(=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}}{\mathrm{n}+1}\right]\) On comparing given equation, we get. \(\frac{\mathrm{kn}}{\mathrm{n}+1}=\frac{1}{4} \frac{\mathrm{n}}{(\mathrm{n}+1)}\) \(\mathrm{k}=\frac{1}{4}\)
AP EAMCET-2012
Sequence and Series
118904
For all \(n \in \mathbb{N}\), if \(\mathbf{1}^2+2^2+3^2+\ldots \ldots .+n^2>x\), then \(\mathbf{x}=\)
1 \(\frac{\mathrm{n}^3}{3}\)
2 \(\frac{\mathrm{n}^3}{2}\)
3 \(\mathrm{n}^3\)
4 \(\frac{n^4}{4}\)
Explanation:
A Given that, \(1^2+2^2+3^2+\ldots . . n^2>x\) \(\sum_{n=1}^n n^2>x\) We know that, \(\sum n^2=\frac{n(n+1)(2 n+1)}{6}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\mathrm{x}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\frac{\mathrm{n}(\mathrm{n})(2 \mathrm{n}) .}{6}\) \(\mathrm{x}=\frac{\mathrm{n} \times \mathrm{n} \times 2 \mathrm{n}}{6}\) \(\mathrm{x}=\frac{\mathrm{n}^3}{3}\)
TS EAMCET-03.05.2019
Sequence and Series
118905
If \(f: R \rightarrow R\) is defined as \(f(x+y)=f(x)+f(y)\), \(\forall x, y \in R\) and \(f(1)=10\), then, \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{f}(\mathrm{r}))^2=\)
D Let, \(\mathrm{T}_{\mathrm{r}} =\mathrm{r}(\mathrm{r}+1)(\mathrm{r}+2)\) \(=\mathrm{r}\left(\mathrm{r}^2+3 \mathrm{r}+2\right)\) \(=\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\) Required sum of \(T_{\mathrm{r}}\) is \(\sigma=\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)\) As we know that \(\sum \mathrm{n}^3=\left[\frac{\mathrm{n}+(\mathrm{n}+1)}{2}\right]^2 \Rightarrow \sum \mathrm{n}^3=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) \(\sum \mathrm{n}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) \(\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)=\) \(\left(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+3 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\) \(=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+\frac{\mathrm{n}(2 \mathrm{n}+1)(\mathrm{n}+1)}{2}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{1}+2(2 \mathrm{n}+1)+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\mathrm{n}^2+\mathrm{n}+4 \mathrm{n}+2+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left(\mathrm{n}^2+5 \mathrm{n}+6\right)\) \(=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\)
AP EAMCET-2015
Sequence and Series
118902
Sum of the first 100 terms of the series \(1+3+7\) \(+\mathbf{1 5}+\mathbf{3 1}+\ldots .\). is
1 \(2^{100}-102\)
2 \(2^{101}-102\)
3 \(2^{102}-103\)
4 \(2^{102}-104\)
Explanation:
B As per give summation, the successive difference are \(2,4,8,16, \ldots\). which are in A.P. here, \(\mathrm{S}=1+3+7+15+31+\ldots \ldots+\mathrm{T}_{100}\) \(=\left(1^2-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\ldots \ldots+\left(2^{100}-1\right)\) \(=\left(2+2^2+2^3+\ldots .+2^{100}\right)-100\) \(=2\left[\frac{2^{100}-1}{2-1}\right]-100\) \(=2^{101}-102\)
J and K CET-2015
Sequence and Series
118903
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots . .(\mathrm{n}\) terms \()=\frac{\mathrm{kn}}{\mathrm{n}+1}\), then \(\mathrm{k}\) is equal to
1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
A Here, given series can be re-written as \(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots .+\frac{1}{2 n}-\frac{1}{2 n+2}\right]\) \(=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}}{\mathrm{n}+1}\right]\) On comparing given equation, we get. \(\frac{\mathrm{kn}}{\mathrm{n}+1}=\frac{1}{4} \frac{\mathrm{n}}{(\mathrm{n}+1)}\) \(\mathrm{k}=\frac{1}{4}\)
AP EAMCET-2012
Sequence and Series
118904
For all \(n \in \mathbb{N}\), if \(\mathbf{1}^2+2^2+3^2+\ldots \ldots .+n^2>x\), then \(\mathbf{x}=\)
1 \(\frac{\mathrm{n}^3}{3}\)
2 \(\frac{\mathrm{n}^3}{2}\)
3 \(\mathrm{n}^3\)
4 \(\frac{n^4}{4}\)
Explanation:
A Given that, \(1^2+2^2+3^2+\ldots . . n^2>x\) \(\sum_{n=1}^n n^2>x\) We know that, \(\sum n^2=\frac{n(n+1)(2 n+1)}{6}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\mathrm{x}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\frac{\mathrm{n}(\mathrm{n})(2 \mathrm{n}) .}{6}\) \(\mathrm{x}=\frac{\mathrm{n} \times \mathrm{n} \times 2 \mathrm{n}}{6}\) \(\mathrm{x}=\frac{\mathrm{n}^3}{3}\)
TS EAMCET-03.05.2019
Sequence and Series
118905
If \(f: R \rightarrow R\) is defined as \(f(x+y)=f(x)+f(y)\), \(\forall x, y \in R\) and \(f(1)=10\), then, \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{f}(\mathrm{r}))^2=\)
D Let, \(\mathrm{T}_{\mathrm{r}} =\mathrm{r}(\mathrm{r}+1)(\mathrm{r}+2)\) \(=\mathrm{r}\left(\mathrm{r}^2+3 \mathrm{r}+2\right)\) \(=\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\) Required sum of \(T_{\mathrm{r}}\) is \(\sigma=\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)\) As we know that \(\sum \mathrm{n}^3=\left[\frac{\mathrm{n}+(\mathrm{n}+1)}{2}\right]^2 \Rightarrow \sum \mathrm{n}^3=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) \(\sum \mathrm{n}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) \(\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{r}^3+3 \mathrm{r}^2+2 \mathrm{r}\right)=\) \(\left(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+3 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\) \(=\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}+\frac{\mathrm{n}(2 \mathrm{n}+1)(\mathrm{n}+1)}{2}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{1}+2(2 \mathrm{n}+1)+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left[\mathrm{n}^2+\mathrm{n}+4 \mathrm{n}+2+4\right]\) \(=\frac{\mathrm{n}(\mathrm{n}+1)}{4}\left(\mathrm{n}^2+5 \mathrm{n}+6\right)\) \(=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\)
AP EAMCET-2015
Sequence and Series
118902
Sum of the first 100 terms of the series \(1+3+7\) \(+\mathbf{1 5}+\mathbf{3 1}+\ldots .\). is
1 \(2^{100}-102\)
2 \(2^{101}-102\)
3 \(2^{102}-103\)
4 \(2^{102}-104\)
Explanation:
B As per give summation, the successive difference are \(2,4,8,16, \ldots\). which are in A.P. here, \(\mathrm{S}=1+3+7+15+31+\ldots \ldots+\mathrm{T}_{100}\) \(=\left(1^2-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\ldots \ldots+\left(2^{100}-1\right)\) \(=\left(2+2^2+2^3+\ldots .+2^{100}\right)-100\) \(=2\left[\frac{2^{100}-1}{2-1}\right]-100\) \(=2^{101}-102\)
J and K CET-2015
Sequence and Series
118903
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots . .(\mathrm{n}\) terms \()=\frac{\mathrm{kn}}{\mathrm{n}+1}\), then \(\mathrm{k}\) is equal to
1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
A Here, given series can be re-written as \(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots .+\frac{1}{2 n}-\frac{1}{2 n+2}\right]\) \(=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right]\) \(=\frac{1}{4}\left[\frac{\mathrm{n}}{\mathrm{n}+1}\right]\) On comparing given equation, we get. \(\frac{\mathrm{kn}}{\mathrm{n}+1}=\frac{1}{4} \frac{\mathrm{n}}{(\mathrm{n}+1)}\) \(\mathrm{k}=\frac{1}{4}\)
AP EAMCET-2012
Sequence and Series
118904
For all \(n \in \mathbb{N}\), if \(\mathbf{1}^2+2^2+3^2+\ldots \ldots .+n^2>x\), then \(\mathbf{x}=\)
1 \(\frac{\mathrm{n}^3}{3}\)
2 \(\frac{\mathrm{n}^3}{2}\)
3 \(\mathrm{n}^3\)
4 \(\frac{n^4}{4}\)
Explanation:
A Given that, \(1^2+2^2+3^2+\ldots . . n^2>x\) \(\sum_{n=1}^n n^2>x\) We know that, \(\sum n^2=\frac{n(n+1)(2 n+1)}{6}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\mathrm{x}\) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}>\frac{\mathrm{n}(\mathrm{n})(2 \mathrm{n}) .}{6}\) \(\mathrm{x}=\frac{\mathrm{n} \times \mathrm{n} \times 2 \mathrm{n}}{6}\) \(\mathrm{x}=\frac{\mathrm{n}^3}{3}\)
TS EAMCET-03.05.2019
Sequence and Series
118905
If \(f: R \rightarrow R\) is defined as \(f(x+y)=f(x)+f(y)\), \(\forall x, y \in R\) and \(f(1)=10\), then, \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{f}(\mathrm{r}))^2=\)