Sequence and Series

$l=\underset{n\to \infty }{lim}{\left(u_n\right)}^{\frac{-4}{n^2}}=\underset{n\to \infty }{lim}{\left\{\left(1+\frac{1}{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2{\left(1+\frac{3^2}{n^2}\right)}^3......{\left(1+\frac{n^2}{n^2}\right)}^n\right\}}^{\frac{-4}{n^2}},$ Taking log on both the sides

$\underset{n\to \infty }{lim}\frac{-4}{n^2}{\displaystyle \sum _{r=1}^{n}rlog\left(1+\frac{r^2}{n^2}\right)}=\underset{n\to \infty }{lim}\frac{-4}{n}{\displaystyle \sum _{r=1}^n\frac{r}{n}log\left(1+{\left(\frac{r}{n}\right)}^2\right)}$

$\therefore logl=-4{\displaystyle \underset{0}{\overset{1}{\int }}xlog\left(1+x^2\right)dx}$

$logl=-2\left[log4-1\right]=-2log\frac{4}{e}=log\frac{e^2}{16}$       

$l=\frac{e^2}{16}$

$S_n=\left\{z\in C:\left|z-3+2i\right|=\frac{n}{4}\right\}$

represents a circle with centre C1 (3, 2) and radius

$r_1=\frac{n}{4}$

Similarly Tn represents circle with centre C2 (2, 3) and radius

$r_1=\frac{1}{n}$

$\sqrt[]{2}<\left|\frac{n}{4}-\frac{1}{n}\right|$

n take infinite values.

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$

$a_{n+2}=2\cdot a_{n+1}-a_{n+1}$

$a_2=2a_1-a_0+1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\ge 2,a_n=\frac{n\left(n-1\right)}{2}$

${\displaystyle \sum _{n=2}^{}\frac{n\left(n-1\right)}{2\cdot 7^n}=\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+......}$

Let S = $\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+.....$

$\frac{\frac{S}{7}=\frac{1}{7^3}+\frac{3}{7^4}+....}{S-\frac{S}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....}$

$\frac{\frac{6}{7}S\cdot \frac{1}{7}=\frac{1}{7^3}+\frac{2}{7^3}+....}{\frac{6}{7}S-\frac{6}{49}S=\frac{1}{7^2}+\frac{1}{7^3}+....}$

$\frac{36}{49}S=\frac{\frac{1}{7^2}}{1-\frac{1}{7}}$

$S=\frac{1}{7^2}*\frac{7}{6}*\frac{49}{36}$

$S=\frac{7}{216}$

Given series $\left\{3*1\right\},\left\{3*2,3*3,3*4\right\},\left\{3*5,3*6,3*7,3*8,3*9\right\}.........$

$\therefore$ 11th set will have 1 + (10)2 = 21 terms

Also up to 10th set total 3 * k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3*101,3*102,......3*121\right\}$ $\therefore$ Sum of elements = 3 * (101 + 102 + ….+121)

$=\frac{3*222*21}{2}=6993.$

Factors of 36 = 2².3².1

Five-digit combinations can be

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$

$d_1=\frac{199-100}{2}\cancel{\in }l$

$d_2=\frac{199-100}{3}=33$

$d_3=\frac{199-100}{4}\cancel{\in }l$

$d_n=\frac{199-100}{i+1}\in l$

$\Rightarrow di=33+11or9$

$\therefore$ sum of common differences = 33 + 11 + 9 = 53

 $S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$

$si{n}^2\theta =co{s}^2\theta$ $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$