Differentiation (Part-38) | S N De

 $\,15.\,\,$ Find the sum of the following series :

$\,(i)\,\,3 \cdot 1^2+4 \cdot 2^2+5 \cdot 3^2+\cdots +(n+2) \cdot n^2 $

Sol. $\,\, t_r=(r+2)r^2=r^3+2r^2$

So, $\,\,\sum_{r=1}^n (r^3+2r^2)\\=\sum_{r=1}^n r^3+2\sum_{r=1}^n r^2\\=\left[\frac{n(n+1)}{2}\right]^2 +2 \cdot \frac{n(n+1)(2n+1)}{6}\\=\frac{n^2(n+1)^2}{4}+\frac 13 n(n+1)(2n+1)\\=\frac{n(n+1)}{12}\left[3n(n+1)+4(2n+1)\right]\\=\frac{n(n+1)}{12} [3n^2+3n+8n+4]\\=\frac{n(n+1)(3n^2+11n+4)}{12}\,\,\text{(ans.)}$

$(ii)\,\,1+3+6+10+15+ \cdots \text{ to n terms}$

Sol. Let $\,S_r=1+3+6+10+\cdots +t_{r} \\ S_r=\,\,\,\,\,\,\,\,\,\,1+3+6+\cdots+t_{r-1}+t_r\\--------------------------\\ \text{Subtracting,}\,\,0=1+2+3+4+\cdots+r-t_r \\ \Rightarrow t_r=\frac{r(r+1)}{2} =\frac {r^2}{2}+\frac r2 \\ \text{So,}\,\,\sum_{r=1}^n t_r=\frac 12 \sum_{r=1}^n r^2+\frac 12\sum_{r=1}^n r \\ \Rightarrow S_n=\frac 12 \cdot \frac{n(n+1)(2n+1)}{6}+\frac 12 \cdot \frac{n(n+1)}{2}\\~~~~~~~~~=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}\\~~~~~~~~~=\frac{n(n+1)}{4}[\frac{2n+1}{3}+1]\\~~~~~~~~~=\frac{n(n+1)}{4} \times \frac{2(n+2)}{3}\\~~~~~~~~~=\frac{n(n+1)(n+2)}{6}\,\,\text{(ans.)}$

$\,(iii)\,\,1+5+12+22+35+\cdots+\text{ to n terms}$

Sol. Let $\, S=1+5+12+22+35+\cdots \\~~~=1+(1+4)+(1+4+7)\\~~~~~~ +(1+4+7+10)+\cdots \\ \therefore t_r=1+4+7+10+\cdots+ \text{ to r terms}\\~~~~~~=\frac r2[2 \times 1+(r-1)\cdot 3] \\ \Rightarrow t_r=\frac r2(3r-1)=\frac{3r^2}{2}- \frac r2 \\ \therefore \sum_{r=1}^n t_r=\frac 32\sum_{r=1}^n r^2-\frac 12 \sum_{r=1}^n r\\ \Rightarrow S_n=\frac 32 \cdot \frac{n(n+1)(2n+1)}{6}-\frac 12 \cdot \frac{n(n+1)}{2}\\~~~~~~~=\frac{n(n+1)}{4}\left[2n+1-1\right]\\~~~~~~~=\frac{n^2(n+1)}{2}\,\,\text{(ans.)}$

$\,16.\,$ The sum of first $\,n\,$ terms of a series is $\,n^2+an+b.\,\,$ Show that $\,b=0\,\,$ and the series is in A.P.

Sol. $\,\,S_n=n^2+an+b \rightarrow(1)$

For $\,\,n=0,\,S_n=0 \\ \therefore b=0 \\ \therefore S_n=n^2+an \\ \text{Now,}\,\, t_n=S_n-S_{n-1}\\~~~~~~~~~~~~~~=n^2+an-[(n-1)^2+a(n-1)]\\~~~~~~~~~~~~~~=n^2+an-n^2+2n-1-an+a\\~~~~~~~~~~~~~~=2n+a-1 \\ \therefore t_{n-1}=2(n-1)+a-1\\~~~~~~~~~~~=2n+a-3 \\ \text{So,}\,t_n-t_{n-1}=2n+a-1-2n-a+3\\~~~~~~~~~~~~~~~~~~~~~~=2 \rightarrow(1)$

So, from $\,(1)\,$ we notice that the difference between two successive terms is constant and hence the series is in A.P. 

$\,17.\,$ If $\,n\,$ is a positive integer, show that,

$(n+1)^2+(n+2)^2+\cdots +4n^2 \\~=\frac{n(2n+1)(7n+1)}{6}$

Sol.$\,(n+1)^2+(n+2)^2+\cdots +4n^2\\=(n+1)^2+(n+2)^2+\cdots+(n+n)^2\\=(n^2+2n+1^2)+(n^2+2.n.2+2^2)+\cdots \\+(n^2+2.n.n+n^2)\\=n.n^2+2n(1+2+\cdots +n)\\+(1^2+2^2+\cdots+n^2)\\=n^3+2n \times \frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}\\=n^3+n^2(n+1)+\frac{n(n+1)(2n+1)}{6}\\=\frac n6[6n^2+6n(n+1)+(n+1)(2n+1)]\\=\frac n6[6n^2+6n^2+6n+2n^2+3n+1]\\=\frac n6[14n^2+9n+1]\\=\frac{n(2n+1)(7n+1)}{6}\,\,\text{(ans.)}$

$\,18.\,$ Find the sum to $\,n\,$ terms of each of the following series :

$\,(i)\,(1)+(1+2)+(1+2+3)+\cdots$

Sol. $\,t_r=1+2+3+4+\cdots \text{to r terms}\\~~~~=\frac{r(r+1)}{2}\\~~~~=\frac{r^2}{2}+\frac r2 \\ \therefore \sum_{r=1}^n t_r=\frac 12\sum_{r=1}^n r^2+\frac 12\sum_{r=1}^n r \\ \Rightarrow S_n=\frac 12\cdot \frac{n(n+1)(2n+1)}{6}+\frac 12 \cdot\frac{n(n+1)}{2}\\~~~~~~~~~~=\frac{n(n+1)}{12}[(2n+1)+3]\\~~~~~~~~~~=\frac{n(n+1)\times 2(n+2)}{12}\\~~~~~~~~~~=\frac{n(n+1)(n+2)}{6}\,\,\text{(ans.)}$

$\,(ii)\,(1^2)+(1^2+2^2)+(1^2+2^2+3^2)+\cdots$

Sol. $\,t_r=1^2+2^2+3^2+\cdots\text{to r terms}\\~~~~=\frac{r(r+1)(2r+1)}{6}\\~~~~=\frac{(r^2+r)(2r+1)}{6}\\~~~~=\frac{2r^3+3r^2+r}{6}\\~~~~=\frac 13 r^3+\frac 12 r^2+\frac 16 r \\ \Rightarrow  \sum_{r=1}^n t_r=\frac 13 \sum_{r=1}^n r^3+\frac 12 \sum_{r=1}^n r^2 \\ ~~~~~~~~~+\frac 16 \sum_{r=1}^n r \\ \Rightarrow S_n=\frac 13\cdot \frac{n^2(n+1)^2}{4}+\frac 12\cdot \frac{n(n+1)(2n+1)}{6} \\ ~~~~~~~~~+\frac 16\cdot \frac{n(n+1)}{2}\\~~~~~~~~~~=\frac{n(n+1)}{12}[n(n+1)+(2n+1)+1]\\~~~~~~~~~~=\frac{n(n+1)}{12}[n^2+n+2n+2]\\~~~~~~~~~~=\frac{n(n+1)}{12}\times [n(n+1)+2(n+1)]\\~~~~~~~~~~=\frac{n(n+1)}{12}\times (n+1)(n+2)\\~~~~~~~~~~=\frac{n(n+1)^2(n+2)}{12}\,\,\text{(ans.)}$

$\,(iii)\,(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+\cdots$

Sol. $\,\,t_r=(2r+1)^3-(2r)^3\\~~~~~=(2r)^3+3.(2r)^2.1+3.(2r).1^2 \\~~~~~~~+1^3-(2r)^3\\~~~~~=12r^2+6r+1 \\ \Rightarrow \sum_{r=1}^n t_r \\ =12 \sum_{r=1}^n r^2+6 \sum_{r=1}^n r+ \sum_{r=1}^n 1\\=12\cdot \frac{n(n+1)(2n+1)}{6}+6\cdot \frac{n(n+1)}{2}+n\\=n[2(n+1)(2n+1)+3(n+1)+1]\\=n[2(2n^2+3n+1)+3n+3+1]\\=n(4n^2+9n+6)\,\,\text{(ans.)}$

$\,(iv)\,(1)+(2+3)+(4+5+6)+\cdots$

Sol. By question, number of terms in first bracket is $\,1\,$, in the second bracket is $\,2\,$, in the third bracket is $\,3\,$, etc. 

Therefore, the number of terms in the $\,n\,-$th bracket will be $\,n\,$.

Let the sum of the given series of $\,n\,$ terms $=S_n$

So, the number of terms in $\,S_n=1+2+3+\cdots +n=\frac{n(n+1)}{2}=p\,\text{(say)}.$

Also, the first term of $\,S_n\,$ is $\,1\,$ and common difference is also $\,1\,$.

$\,\therefore S_n=\frac{p}{2}[2 \times 1+(p-1)\times 1]\\~~~~~~~~~=\frac p2(p+1)\\~~~~~~~~~=\frac 12 (p^2+p)\\~~~~~~~~~=\frac 12\left[\frac{n^2(n+1)^2}{4}+\frac{n(n+1)}{2}\right]\\~~~~~~~~~=\cdots \\~~~~~~~~~=\frac{n(n+1)(n^2+n+2)}{8}\,\,\text{(ans.)}$