Differentiation (Part-38) | S N De

 $\,9.\,$ Write the series $\,\,\sum_{r=1}^n\frac{2r+1}{r^2+1}\,\,$ in expanded form.

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

So, $\,u_1=\frac{2.1+1}{1^2+1}=\frac 32, \\ u_2=\frac{2.2+1}{2^2+1}=1,\\ u_3=\frac{2.3+1}{3^2+1}=\frac{7}{10}, \\ u_4=\frac{2.4+1}{4^2+1}=\frac{9}{17}, \\ \vdots \\ u_n=\frac{2n+1}{n^2+1}.$

Hence, the series in expanded form can be written as : 

$\,\,\frac 32+1+\frac{7}{10}+\frac{9}{17}+\cdots+\frac{2n+1}{n^2+1}.$

$\,10.\,$ If $\,S_n=u_1+u_2+\cdots+u_n=n^2+2n,\,$ find the first four terms of the series.

Sol. $\,u_n\,\,[\text{=n-th term of the series}]\\=S_{n}-S_{n-1}\\=(n^2+2n)-[(n-1)^2+2(n-1)]\\=n^2+2n-(n-1)(n-1+2)\\=n^2+2n-(n-1)(n+1)\\=n^2+2n-(n^2-1)\\=2n+1$

So, $\,u_1=2.1+1=3,\\ u_2=2.2+1=5,\\u_3=2.3+1=7,\\ u_4=2.4+1=9.$

$\,11.\,$ If the $\,r^{th}\,$ term of the sequence $\,\{u_n\}\,$ is $\,u_r=(-1)^{r-1}.3^{3-r},\,$ find the first five terms of the sequence ; also find the corresponding series.

Sol. $\,u_1=(-1)^{1-1}.3^{3-1}=3^2=9,\\ u_2=(-1)^{2-1}.3^{3-2}=-3,\\ u_3=(-1)^{3-1}.3^{3-3}=1,\\ u_4=(-1)^{4-1}.3^{3-4}=-\frac 13,\\ u_5=(-1)^{5-1}.3^{3-5}=\frac 19.$

The corresponding series is :$\,9+(-3)+1+(-1/3)+1/9.$ 

$\,12.\,\,$ The sum of first $\,r\,$ terms of a series is $\,ar^2+br \,\,;\,\,$ Find the first and $\,12^{th}\,$ terms of the series.

Sol. The sum of first $\,r\,\,$ of the series : $\,S_r=ar^2+br ; $

So, the first term : $\,S_1=u_1=a.1^2+b.1=a+b.$

The $\,r^{th}\,$ term of the series is : 

$u_r\\=S_r-S_{r-1}\\=ar^2+br-[a(r-1)^2+b(r-1)]\\=a[r^2-(r-1)^2]+b[r-(r-1)]\\=a[(r+r-1)(r-r+1)]+b\\=a(2r-1)+b \\ \therefore u_{12}=a(2 \times 12-1)+b=23a+b\,\,\text{(ans.)}$

$\,13.\,$ For the sequence $\,\{u_n\}\,\,$ if $\,u_1=-2\,$ and $\,\,u_{r+1}=u_r+r+2,\,$ for all natural number $\,r\,$; find the $\, 10^{th}\,$ term of the sequence.

Sol. We have, $\,\,u_{r+1}=u_r+r+2\,\rightarrow(1).$ 

Putting $\,\,r=9,8,7,\cdots ,2,1\,$ we get from $\,(1)\,$, 

$\, u_{10}=u_9+9+2,\\ u_9=u_8+8+2,\\ u_8=u_7+7+2, \\ \vdots \\ u_2=u_1+1+2 \\ ------------------\\ \text{Adding ,}\, (u_{10}+u_9+\cdots+u_2)=(u_9+u_8 \\ +\cdots +u_1)+(9+8+7+\cdots+1)+ 2\times 9 \\ \Rightarrow u_{10}=u_1+\frac{9(9+1)}{2}+18 \\ \Rightarrow u_{10}=-2+ 9 \times 5+18=61\,\,\text{(ans)} $

$\,14.\,$ For the sequence $\,\{u_n\}\,$ if $\,u_1=\frac 14\,$ and $\,u_{n+1}=\frac{u_n}{2+u_n},\,$ find the value of $\,\frac{1}{u_{50}}.$

Sol. We have, $\,u_{n+1}=\frac{u_n}{2+u_n} \\ \Rightarrow \frac{1}{u_{n+1}}=\frac{2+u_n}{u_n}=(2/u_n)+1 \rightarrow(1)$

Putting $\,\,n=1,2,3,\cdots ,49\,\,$ in $\,(1),\,\,$ we get,

$\,1/u_2=(2/u_1)+1 \\ 1/u_3=(2/u_2)+1 \\ ~~~~~~~~=2(2/u_1+1)+1\\ ~~~~~~~~=\frac{2^2}{u_1}+2+1 \\ 1/u_4=2/u_3+1\\ ~~~~~~~~=2(\frac{2^2}{u_1}+2+1)+1\\ ~~~~~~~~=\frac{2^3}{u_1}+2^2+2+1\\ \vdots \\ \frac{1}{u_{50}}=\frac{2}{u_{49}}+1\\ ~~~~~=\frac{2^{49}}{u_{1}}+2^{48}+2^{47}+\cdots+2+1\\ ~~~~~=4\times 2^{49}+2^{48}+\cdots +2+1\\ ~~~~~= 4 \times 2^{49}+\frac{2^{49}-1}{2-1}\\ ~~~~~=4 \times 2^{49}+ 2^{49}-1\\ ~~~~~=5^{49}-1\,\,\text{(ans.)}$ 

$\,15.\,$ Find the least value of $\,n\,$, for which the $\,n^{th}\,$ term $\,a_n\,$of the sequence  given by $\, a_n=n^3-n^2-5n-3\,$ is non-negative.

Sol. We have, $\, a_n=n^3-n^2-5n-3\,\rightarrow(1)$.

Putting $\,n=1,2,3 \cdots$ in $\,(1)\,$ we get,

$\,a_1=1^3-1^2-5\times 1-3 =-8,\\ a_2=2^3-2^2-5\times 2-3=-9,\\a_3=3^3-3^2-5\times 3-3=0.$

Hence  the least value of $\,n\,$, for which the given  sequence  is non-negative is $\,3.$

$\,16.\,$ A sequence whose $\, n^{th}\,$ is given by $\,a_n=2n^2+pn-3p^2\,$ is such that $\,a_3=0.$ Determine the sequence explicitly and then  show that $\,a_n >0 \,\,\,\forall n >4.$ 

Sol. $\,a_n=2n^2+pn-3p^2\,\,\,\text{(given)} \\ \Rightarrow a_3=2.3^2+3p-3p^2 \\ \Rightarrow 0=18+3p-3p^2 \\ \Rightarrow 3(p^2-p-6)=0 \\ \Rightarrow p^2-p-6=0 \\ \Rightarrow p^2-3p+2p-6=0 \\ \Rightarrow p(p-3)+2(p-3)=0 \\ \Rightarrow (p-3)(p+2)=0 \\ \therefore p=3,-2 \rightarrow (1)$

Hence, from $\,(1)\,$, we get,  $\,a_n=2n^2+3n-3.3^2\,\,\text{(for p=3)}\\ \Rightarrow a_n=2n^2+3n-27 \rightarrow(2) \\ \text{or,}\,\, a_n=2n^2-2n-3(-2)^2 \\ \Rightarrow a_n=2n^2-2n-12,\,\,\forall n \in N \rightarrow(3)$

Now from $\,(2),\,(3),\,$ we get, $\,a_1=2+3-27=-22 \\ \text{or,}\,\,a_1=2-2-12=-12$

Again, from $\,(2),\,(3),\,$ we get, $\,a_2=2.2^2+3.2-27=-13 \\ \text{or,}\, a_2=2.2^2-2.2-12=-8.$

Similarly, we see  [from $\,(2),\,(3),\,$] $\,a_3=0,0.$

But $\,a_4=2.4^2-3.4-27=17 \\ \text{or,}\,\,\,a_4=2.4^2-2.4-12=12$

Similarly, we can show that for $\,n >4,\, a_n >0.$