Sequences and Series

$f\left(x\right)=\frac{5^x}{5^x+5}$

$f\left(2-x\right)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{25/5^x}{\frac{25}{5^x}+5}=\frac{25}{25+5^{x+1}}=\frac{5}{5+5^x}$

$f\left(x\right)+f\left(2-x\right)=1$

$S={\displaystyle \sum _{k=1}^{39}f\left(\frac{k}{20}\right).....\left(i\right)}$

$S={\displaystyle \sum _{k=1}^{39}f}\left(2-\frac{k}{20}\right)..........\left(ii\right)$

(i) + (ii) 2S = ${\displaystyle \sum _{k=1}^{39}\left(f\left(\frac{k}{20}\right)+f\left(2-\frac{k}{20}\right)\right)}$

$S=\frac{39}{2}$

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

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

  $S_{10}=530\Rightarrow 5\left[2a+9d\right]=530$

2a + 9d = 106 . (i)

$S_5=140\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$              

a + 2d = 28 . (ii)

Solving (i) & (ii) a = 88 d = 10

$\therefore S_{20}-S_6=14a+175d=1862$            

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$

$S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}..........\infty$

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^2}+\frac{13}{5^4}+------\infty$

$Let\frac{4S-7}{5}=t$

$\frac{4t}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$t=\frac{1}{8}$

$\frac{4S-7}{5}=\frac{1}{8}$

$S=\frac{61}{32}$

$\therefore 160S=305$

$?\frac{3}{1^2*2^2}+\frac{5}{2^2*3^2}+\frac{7}{3^2*4^2}+........$

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

$S_{10}={\displaystyle \sum _{r=1}^{10}{t}_r=}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}\right)=1-\frac{1}{11^2}=\frac{120}{121}}$

  $\underset{x\to \infty }{lim}\left(\sqrt[]{x^2-x+1}-ax\right)=b$

$\underset{x\to \infty }{lim}\left|\frac{x^2-x+1-a^2{x}^2}{\sqrt[]{x^2-x+1}+ax}\right|=b$

For existence of limit 1 – a² = 0 i.e. a = 1 only

$\underset{x\to \infty }{lim}\frac{1-x}{\sqrt[]{x^2-x+1}+x}=b$

$\Rightarrow b=\frac{-1}{2}$

So, (a, b) =   $\left(1,-\frac{1}{2}\right)$

$\frac{a}{1-r}=15,\frac{a^2}{1-r^2}=150$

$\Rightarrow r=\frac{1}{5},a=12\Rightarrow \frac{a{r}^2}{1-r^2}=\frac{1}{2}$

$S-\frac{1}{x-1}=\frac{-2^{101}}{{\left(x^{2^{100}}\right)}^2-1}$

For x = 2, S = 1 $\frac{2^{101}}{4^{101}-1}$