Sequences and Series

$C_2=a_2+b_2=\left(a_1-3\right)+2b_1=12$

$a_1+2b_1=15.............\left(i\right)$

$a_1+4b_1=19............\left(ii\right)$              

Solving (i) & (ii), b₁ = 2, a₁ = 11

$=5\left[22+9\left(-3\right)\right]+2\left(\frac{2^{10}-1}{2-1}\right)$

$=2^{11}-27=2^6*2^5-27=2021$               

$1.3^0+2.3^1+3.3^2+....+10.3^9$  

${\displaystyle \sum _{n=0}^{10}{x}^n=1+x+x^2+.....+x^{10}=\frac{1.\left(x^{11}-1\right)}{x-1}}$                

For n = 3, S =  $\frac{22.3^{10}-\left(3^{11}-1\right)}{4}$  

$=\frac{3^{10}\left(22-3\right)+1}{4}=\frac{19.3^{10}+1}{4}$                

$\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow d=4a$  

110 < a₁₅ < 120

110 < a + 14d < 120

110 < 57a < 120

->a = 2, d = 8

  $T_r=\frac{r}{{\left(2r^2\right)}^2+1}$

$=\frac{r}{{\left(2r^2+1\right)}^2-{\left(2r\right)}^2}$

$=\frac{1}{4}\frac{4r}{\left(2r^2+2r+1\right)\left(2r^2-2r+1\right)}$

$S_{10}=\frac{1}{4}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{\left(2r^2-2r+1\right)}-\frac{1}{\left(2r^2+2r+1\right)}\right)}$

$\Rightarrow S_{10}=\frac{1}{4}.\frac{220}{221}=\frac{55}{221}=\frac{m}{n}$

$\therefore m+m=276$

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP

$a_{n+2}{a}_{n+1}-a_{n+1}{a}_n=2$

Series will satisfy

$a_1{a}_2,a_2{a}_3,a_3{a}_4,......a_4{a}_5$

$\begin{array}{l}1.22.22.32.4\\ \frac{a_n+\frac{1}{a_{n+1}}}{a_{n+2}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}\end{array}$

$=1-\frac{1}{2\left(r+1\right)}=\frac{2r+1}{2\left(r+1\right)}$

Now, proof = $\frac{30}{11}\frac{\left(2r+1\right)}{2\left(r+1\right)}$ $\frac{}{}\frac{}{}$

r = 1

$=\frac{\left(1.3.5......61\right)}{2^{30}\left(2.3......31\right)}$

$\frac{\overline{)61}}{2^{60}.\overline{)31}.\overline{)30}}=\alpha =-60$

${\displaystyle \sum _{n=1}^{21}\frac{3}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}{\displaystyle \sum _{n=1}^{21}\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}\frac{4n+3-2}{3\left(4n+3\right)}=\frac{n}{4n+3}$

for n = 21

$S_{21}=\frac{21}{84+3}=\frac{21}{87}=\frac{7}{29}$

x² – x – 4 = 0

  $P_n={\alpha }^n-{\beta }^n$                    

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n = aⁿ - bⁿ   

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$                  

 $?a_n={\displaystyle {\int }_{-1}^0\left(1+\frac{x}{2}+\frac{x^2}{2}+....+\frac{x^{n-1}}{n}\right)dx}$

$={\left[x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+.....+\frac{x^n}{n^2}\right]}_{-1}^n$

$a_n=\frac{n+1}{1^2}+\frac{n^2-1}{2^2}=\frac{n^3+1}{3^2}+\frac{n^4-1}{4^2}+.....+\frac{n^n+{\left(-1\right)}^{n+1}}{n^2}$

Here a₁ = 2, $a_2=\frac{2+1}{1}+\frac{2^2-1}{2}=3+\frac{3}{2}=\frac{9}{2}$

$a_4=5+\frac{15}{4}+\frac{65}{9}+\frac{255}{16}>31$

$\therefore$ The required set is {2, 3}

$?a_n\in \left(2,30\right)$

$\therefore$ Sum of elements = 5.

a₁ = b₁ = 1

$a_n=a_{n-1}+2\left(forn\ge 2\right)b_n=a_n=b_{n-1}$          

Similarly for others

${\displaystyle {\sum }_{n=1}^{11}{a}_n{b}_n={\displaystyle {\sum }_{n=1}^{15}\left(2n-1\right)n^2={\displaystyle {\sum }_{n=1}^{15}2n^3-{\displaystyle {\sum }_{n=1}^{15}{n}^2}}}}$      

$=2{\left[\frac{15*16}{2}\right]}^2-\left[\frac{15*16*31}{6}\right]=27560$