Differentiation (Part-38) | S N De

 

$\,18(v)\,\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\cdots$

Sol. $\,t_r=\frac{1}{(3r-2)(3r+1)}=\frac 13\left[\frac{1}{3r-2}-\frac{1}{3r+1}\right]$

Now, $\,\,t_1+t_2+t_3+\cdots +t_n\\=\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\cdots+\frac{1}{(3n-2)(3n+1)}\\=\frac 13\left[\frac 11-\frac 14+\frac 14-\frac 17+\frac 17-\frac{1}{10}\\+\cdots +\frac{1}{3n-2}-\frac{1}{3n+1}\right]\\=\frac 13 \left(1-\frac{1}{3n+1}\right)\\=\frac{n}{3n+1}\,\,\text{(ans.)}$

$\,(vi)\,\,\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\cdots$

Sol. $\,t_r=\frac{1}{(3r-1)(3r+2)}=\frac 13\left(\frac{1}{3r-1}-\frac{1}{3r+2}\right)$

So, $\,\,t_1+t_2+t_3+\cdots +t_n\\=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\cdots+ \frac{1}{(3n-1)(3n+2)}\\=\frac 13 \left[\frac{1}{2}-\frac 15+\frac 15-\frac 18+\frac 18-\frac{1}{11}\\+\cdots+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\\=\frac 13 \left(\frac 12-\frac{1}{3n+2}\right)\\=\frac 13 \times \frac{3n+2-2}{2(3n+2)}\\=\frac{n}{2(3n+2)}\,\,\text{(ans.)}$

$\,(vii)\,\, n\cdot 1+(n-1) \cdot 2+ (n-3) \cdot 4 +\cdots $

Sol. Here, $\,t_r=[n-(r-1)] \cdot r=nr-r^2+r$

So, $\,t_1+t_2+t_3+\cdots +t_n\\=(n \cdot 1 -1^2+1)+(n \cdot 2-2^2+2) \\ +(n \cdot 3-3^2+3) +\cdots + (nr -r^2+r)\\=n(1+2+3+\cdots +n)-(1^2+2^2+3^2 \\ +\cdots +n^2)+(1+2+3+\cdots +n)\\=n \times \frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\=\frac{n(n+1)}{2}\left[n-\frac{2n+1}{3}+1\right]\\=\frac{n(n+1)}{2}\left[\frac{3n-2n-1+3}{3}\right]\\=\frac{n(n+1)}{2} \times \frac{n+2}{3}\\=\frac 16 n(n+1)(n+2)\,\,\text{(ans.)}$

$\,19.\,$ The sum of first $\,P\,$ terms of an A.P. is equal to the sum of its first $\,Q\,$ terms . Show that the sum of its first $\,(P+Q)\,$ terms is zero.

Sol. By question, $\,S_P=S_Q \\ \Rightarrow \frac P2 [2a+(P-1)d]=\frac Q2[2a+(Q-1)d] \\ \Rightarrow P[2a+(P-1)d]=Q[2a+(Q-1)d] \\ \Rightarrow 2aP+P(P-1)d=2aQ+Q(Q-1)d \\ \Rightarrow 2a(P-Q)+d[P(P-1)\\ -Q(Q-1)]=0 \\ \Rightarrow 2a(P-Q)+d[(P^2-Q^2)-(P-Q)]=0 \\ \Rightarrow 2a(P-Q)+d[(P+Q)(P-Q)\\ -(P-Q)]=0 \\ \Rightarrow 2a(P-Q)+d(P-Q)[P+Q-1]=0 \\ \Rightarrow (P-Q)[2a+d(P+Q-1)]=0 \\ \Rightarrow 2a+d(P+Q-1)=0 \,\,[\because P \neq Q] \rightarrow(1)$

Now, $\,S_{P+Q}\\=\frac{P+Q}{2}[2a+(P+Q-1)d]\\=\frac{P+Q}{2} \times 0\,\,[\text{By (1)}]\\=0\,\,\,\text{(showed)}$ 

Note : Here, $\,a\,$ denotes first term of A.P. and $\,d\,$ denotes the common difference of the A.P.

$\,20.\,$ A series is given by in the form : $\,(1)+(2+3+4)+(5+6+7+8+9)+\cdots$ Find the sum of the numbers in the $\,r\,$-th bracket.

Sol. We compute the first number in each bracket. The first numbers in each bracket are given by $\,\,\,\,1,2,5,10,\cdots ,t_n\,\,$ where $\,t_n\,$ denotes the first term in the $\,n\,$-th bracket. 

$\,S_n=1+2+5+10+\cdots+t_n \\ S_n=\quad \quad 1+2+5+\cdots +t_{n-1}+t_n \\ ------------------------\\ \text{Subtracting,}\, 0=1+1+3+5+ \\ \cdots \text{to (n-1) terms}-t_n \\ \Rightarrow t_n= 1+(1+3+5+\cdots \text{to (n-1) terms})\\ \Rightarrow t_n=1+\frac{n-1}{2}[2 \times 1+(n-1-1)\times 2] \\ \Rightarrow t_n=1+\frac{n-1}{2}[2+2(n-2)]\\ \Rightarrow t_n=1+\frac{n-1}{2}(2n-2)\\ \Rightarrow t_n=1+(n-1)(n-1) \\ \Rightarrow t_n=n^2-2n+2$ 

So, the first term of $\,n\,$-th bracket is : $\,t_n=n^2-2n+2.$ 

Let $\,l\,$ be the last term of $\,n\,$-th bracket. We notice that the number of terms in each bracket is $\,\,(2n-1).$

So, $\, l=t_n+(2n-1-1)\times 1\,\,[*] \\ \Rightarrow l=n^2-2n+2+2n-2 \\ \Rightarrow l=n^2.$

Hence, the sum of numbers in the $\,n\,$-th bracket is : 

$S'=\frac{2n-1}{2}[t_n+l]\\ \therefore S'=\frac{2n-1}{2}(n^2-2n+2+n^2) \\ \Rightarrow S'=(2n-1)(n^2-n+1)\\ \Rightarrow S'=2n^3-3n^2+3n-1 \\ \Rightarrow S'=n^3+(n^3-3n^2+3n-1) \\ \Rightarrow S'=n^3+(n-1)^3.$

So, the sum of the numbers in the $\,r\,$-th bracket is : $\,r^3+(r-1)^3\,\,\text{(ans.)}$

Note [*] : Here we have used the formula : 

$\,\,t_n=a+(n-1)d,\,\,$ where $\,t_n\,$ denotes the $\,n\,$-th term, $\,a\,$ being the first term and $\,d\,$ is the common difference.

$\,21.\,$ The first and last terms of an A.P. are $\,a\,$ and $\,l\,$ respectively. If $\,S\,$ be the sum of all terms then show that the common difference is : $\,\frac{l^2-a^2}{2S-(l+a)}.$

Sol. Let the number of terms of the A.P. is $\,n\,\,$ and the common difference is $\,d.$ 

Now, $\,S=\frac n2(a+l),\,\,\,a,l \,\,\text{being first and last term.} \\ \Rightarrow n=\frac{2S}{a+l}$

Again, $\,S=\frac{n}{2}[2a+(n-1)d] \\ \Rightarrow 2S=\frac{2S}{a+l}\left[2a+\left(\frac{2S}{a+l}-1\right)d\right] \\ \Rightarrow a+l=2a+\frac{2S-(a+l)}{(a+l)}\times d \\ \Rightarrow (a+l)^2=2a(a+l)+[2S-(a+l)]d \\ \Rightarrow (a+l)(a+l-2a)=[2S-(a+l)]d \\ \Rightarrow (l+a)(l-a)=[2S-(a+l)]d \\ \Rightarrow d=\frac{l^2-a^2}{2S-(l+a)}\,\,\text{(showed)}$

$\,22.\,$ If the first, second and last terms of an A.P. be $\,a,b,\,c\,$ respectively, show that the sum of all terms of the A.P. is $\,\,\frac{(a+c)(b+c-2a)}{2(b-a)}.$

Sol.  The common difference of the given A.P., $\,d=b-a\rightarrow(1)$

Now, $\,a+(n-1)d=c \\ \Rightarrow a+(n-1)(b-a)=c\,\,\text{[By (1)]} \\ \Rightarrow n-1=\frac{c-a}{b-a} \\ \Rightarrow n=\frac{c-a}{b-a}+1 \\ \Rightarrow n=\frac{c-a+b-a}{b-a}=\frac{b+c-2a}{b-a}\\ \therefore S=\frac n2(a+c)\\ \Rightarrow S=\frac{a+c}{2} \times \frac{b+c-2a}{b-a} \\ \Rightarrow S=\frac{(a+c)(b+c-2a)}{2(b-a)}.$