Differentiation (Part-38) | S N De

$\,1.\,$ Write the first five terms of the sequence defined by : $\,(i)\,u_n=\frac{n}{n+1}\,$and also find the corresponding series .

Sol. Putting $\,n=1,2,3,4,5\,$ we get the first five terms of the sequence which are

 $\,\,u_1=\frac{1}{1+1}=\frac 12,\\ u_2=\frac{2}{2+1}=\frac 23, \\ u_3=\frac{3}{3+1}=\frac 34,\\ u_4=\frac{4}{4+1}=\frac 45,\\  u_5=\frac{5}{5+1}=\frac 56.$ 

The corresponding series is given by : $\,u_1+u_2+u_3+u_4+ \cdots\\=\frac 12+\frac 23+\frac 34+\frac 45+ \cdots$

$\,1.\,$ Write the first five terms of the sequence defined by : $\,(ii)\,u_n=(-1)^n\frac{n}{3n+1}$

Sol. Putting $\,n=1,2,3,4,5\,$ we get the first five terms of the sequence which are $\,\,u_1=(-1)^1\frac{1}{3.1+1}=-\frac 14,\\  u_2=(-1)^2\frac{2}{3.2+1}=\frac 27,\\ u_3=(-1)^3\frac{3}{3.3+1}=-\frac{3}{10},\\u_4=(-1)^4\frac{4}{3.4+1}=\frac{4}{13},\\ u_5=(-1)^5\frac{5}{3.5+1}=-\frac{5}{16}.$

The corresponding series is given by : $\,u_1+u_2+u_3+u_4+ \cdots\\=\left(-\frac 14\right)+\frac 27+\left(-\frac {3}{10}\right)+\frac{4}{13}+ \cdots$

$\,1.\,$ Write the first five terms of the sequence defined by : $\,(iii)\,u_n=2n^2-3n$

Sol. Putting $\,n=1,2,3,4,5\,$ we get the first five terms of the sequence which are 

$\,\,u_1=2.1^2-3.1=-1, \\ u_2=2.2^2-3.2=8-6=2,\\ u_3=2.3^2-3.3=18-9=9,\\ u_4=2.4^2-3.4=32-12=20,\\u_5=2.5^2-3.5=50-15=35.$

$\,2.\,$ Find the $\,\,6^{th} \,$ term  and $\,r^{th}\,$ terms of the sequence $\,\,\{1,8,27,64,\cdots\}$

Sol.  Let $\,u_n\,$ be the $\,n^{th}\,$ term of the given sequence so that $\,u_1=1=1^3,\,u_2=8=2^3,\,u_3=27=3^3, \\ u_4=64=4^3,\cdots.$

So, following the pattern we can say $\,u_n=n^3 \rightarrow(1)$

Hence , from $\,(1)\,$ we can say $\,u_6=6^3=216,\, u_r=r^3.$

$\,3(i)\,$ Prepare a series of the first $\,n^{th}\,$ terms of the sequence $\,\,\{\frac 15,\frac 17,\frac 19,\frac{1}{11},\cdots\}$

Sol.  Let $\,u_n\,$ be the $\,n^{th}\,$ term of the given sequence so that 

$\,u_1=\frac 15=\frac{1}{2.1+3},\\ u_2=\frac 17=\frac{1}{2.2+3},\\u_3=\frac 19=\frac{1}{2.3+3},\\u_4=\frac{1}{11}=\frac{1}{2.4+3}, \\ \vdots \\u_n=\frac{1}{2n+3}$

Hence, the series is given by $\,u_1+u_2+u_3+\cdots +u_n\\=\frac 15+\frac 17+\frac 19+\cdots +\frac{1}{2n+3}$

$\,3(ii)\,$ Find the $\,11^{th}\,\,$ term of the sequence $\,\{\frac 12,\frac 13,\frac 15,\frac 18,\frac{1}{12},\cdots\}.$

Sol.  Let $\,S_n=2+3+5+8+12+\cdots +u_n \rightarrow(1)\\ S_n=~~~~~~~2+3+5+8+\cdots+u_{n-1}+u_n \\ ~~~~~~~~~~~~~~~~~~~\rightarrow(2)$

Note : here, $\,u_n\,$ denotes the $\,n^{th}\,$ term of the sequence $\,\,\{2,3,5,8,12,\cdots\}.$

Subtracting $\,(2)\,$ from $\,(1),\,$ we get 

$0=2+[(3-2)+(5-3)+(8-5) \\ +\cdots+\text{upto (n-1)terms}]-u_n \\ \Rightarrow u_n=2+[1+2+3+\cdots \text{upto (n-1)terms}]\\ \Rightarrow u_n=2+\frac{(n-1)(n-1+1)}{2}\\ \Rightarrow u_n=2+\frac{n(n-1)}{2} \\ \therefore u_{11}=2+\frac{11\times 10}{2}=57.$

Hence, the $\,11^{th}\,\,$ term of the sequence is $\,\,\frac{1}{57}\,\,\text{(ans.)}$

$\,4.\,$ Find the $\,25^{th}\,$ and $\,50^{th}\,$ term of the sequence whose $\,n\,$-th term is given by 

$u_n=\,\begin{cases} \frac{n}{n+1},\,\text{when n is odd}\\ \frac{n+1}{n+2},\,\,\text{when n is even} \end{cases}$

$\,4.\,$ Find the $\,25^{th}\,$ and $\,50^{th}\,$ term of the sequence whose $\,n\,$-th term is given by 

$u_n=\,\begin{cases} \frac{n}{n+1},\,\text{when n is odd}\\ \frac{n+1}{n+2},\,\,\text{when n is even} \end{cases}$

Sol.  Since $\,n=25\text{(odd)},\,$ so by definition of $\,u_n$

$\,u_{25}=\frac{25}{25+1}=\frac{25}{26}$

Again, since $\,n=50\text{(even)},\,$ so by definition of $\,u_n$

$\,u_{50}=\frac{50+1}{50+2}=\frac{51}{52}$

$\,5.\,$ Write the series $\,\sum_{r=1}^n \frac{r^2+1}{2r^2-1}\,\,$ in expanded form. 

Sol. Clearly, here the n-th term=$\,u_n=\frac{n^2+1}{2n^2-1}$.

So, $\,u_1=\frac{1^2+1}{2.1^2-1}=\frac{2}{2-1}=2,\\ u_2=\frac{2^2+1}{2.2^2-1}=\frac{5}{8-1}=\frac 57,\\ u_3=\frac{3^2+1}{2.3^2-1}=\frac{10}{17},\\ u_4=\frac{4^2+1}{2.4^2-1}=\frac{17}{31}, \\ \vdots \\ u_n=\frac{n^2+1}{2n^2-1}$

Hence, the series in expanded form is given by : 

$\,\, 2+\frac 57+\frac{10}{17}+\frac{17}{31}+\cdots+ \frac{n^2+1}{2n^2-1}.$

$\,6.\,$ Determine the first five terms of the sequence defined by , $\,u_1=-2,\,u_2=2\,$ and $\,u_n=\frac{n}{n-2}u_{n-1},\,\,n >2.$

Sol. $\,u_3=\frac{3}{3-2}u_{3-1}=3u_2=3 \times 2=6, \\ u_4=\frac{4}{4-2}u_{4-1}=2u_3=2\times 6=12,\\u_5=\frac{5}{5-2}u_{5-1}=\frac 53 u_4=\frac 53\times 12=20.$

Hence, the first five terms of the sequence are given by $\,\,-2,\,2,\,6,\,12,\,20.$

$\,7.\,$ Write the series $\,\,\frac 32+1+\frac{7}{10}+\frac{9}{17}+\cdots+\frac{2r+1}{r^2+1}\,\,$ in sigma notation.

Sol. The $\,n^{th}\,$ term of the series is : $\,u_n=\frac{2n+1}{n^2+1}$

Hence , the series is given by :

$\,\sum_{r=1}^n u_r \\=\sum_{r=1}^n \left(\frac{2r+1}{r^2+1}\right)\,\,\text{(ans.)}$

$\,8.\,$ Determine the first five terms of the sequence defined by, $\,u_1=4,\,$ and $\,u_n=3u_{n-1}+2\,$ for $\,\,n >2.\,$ Also find the series of first five terms of the sequence.

Sol. $\,u_1=4,\\ u_2=3u_1+2=3\times4+2=14,\\ u_3=3u_2+2=3\times14+2=44,\\ u_4=3u_3+2=3\times44+2=134,\\u_5=3u_4+2=3\times134+2=404.$

Hence, the sequence is given by : $\,4,\,14,\,44,\,134,\,404\,\,$ and the series of first five terms of the sequence is $\,\,4+14+44+134+404.$