Sequences and Series

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$ …… (i)

Now,  ${\left(-2\right)}^3\equiv -1mod\left(7\right)$

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$ ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$

$\therefore$ Remainder = 5

Given ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{rr}& \frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}=\sum _{r=0}^{20}??\frac{a_r{2}^r}{x^r}\\ & \Rightarrow 2^{10}\sum _{r=0}^{20}??a_r{x}^r=\sum _{r=0}^{20}??a_r{2}^r{x}^{(20-r)}\left(\text{from}\right(i\left)\right)\end{array}$

now, comparing coefficient of $x^7$ from both sides

(take $r=7$ in L.H.S. and $r=13$ in R.H.S.)

$2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=2^3=8$

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35*11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

$x\frac{dy}{dx}-y=x^2(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x),x>0$

$\frac{dy}{dx}-\frac{y}{x}=x(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x)\Rightarrow \frac{dy}{dx}+Py=Q$

So, I.F. $=e^{\int -\frac{1}{x}dx}\frac{1}{\left|x\right|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}\left(x\right(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x\left)\right)dx$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}+\mathrm{C}$

$?\mathrm{y}\left(\pi \right)=\pi \Rightarrow \mathrm{C}=1$

So, $y=x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+x\Rightarrow (y{)}_{\pi /2}=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Also, $\frac{dy}{dx}=x^2\mathrm{c}\mathrm{o}\mathrm{s}?x+2x\mathrm{s}\mathrm{i}\mathrm{n}?x+1$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+4x\mathrm{c}\mathrm{o}\mathrm{s}?x+2\mathrm{s}\mathrm{i}\mathrm{n}?x$

6. Here, a_n=n $\frac{x^2+5}{4}$

Putting n=1,2,3,4,5 we get,

$a_1=1*\frac{\left(1^2+5\right)}{4}=1*\frac{6}{4}=\frac{3}{2}$

$a_2=2*\left(\frac{2^2+5}{4}\right)=2*\frac{(4+5)}{4}=\frac{9}{2}$

$a_3=3*\frac{\left(3^2+5\right)}{4}=3*\frac{(9+5)}{4}=3*\frac{14}{4}=\frac{21}{2}$

$a_4=4*\frac{\left(4^2+5\right)}{4}=4*\frac{(16+5)}{4}=4*\frac{21}{4}=21$

$a_5=5*\frac{\left(5^2+5\right)}{4}=5*\frac{(25+5)}{4}=5*\frac{30}{4}=\frac{75}{2}$

Hence, the first five terms are $\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$ .

3. Here a_n=2ⁿ

Substituting n=1,2,3,4,5 we get,

a₁=2¹=2

a₂=2²=4

a₃=2³=8

a₄=a⁴=16

a₅=2⁵=32.

Hence the first five terns are 2,4,16,32 and 64.

2. Here, a₁= $\frac{n}{n+1}$

Substituting n=1,2,3,4,5 we get,

$a_1=\frac{1}{1+1}=\frac{1}{2}$

$a_2=\frac{2}{2+1}=\frac{2}{3}$

$a_3=\frac{3}{3+1}=\frac{3}{4}$

$a_4=\frac{4}{4+1}=\frac{4}{5}$

$a_5=\frac{5}{5+1}=\frac{5}{6}$ .

Hence the first five terns are $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}$ and $\frac{5}{6}$ .

105. Given,

Cost of machine =? 15625

depreciation rate = 20 % each year.

We have,

Depreciated value after 1st year =? 15625 - 20 % of 15625

=?  $15625-\frac{20}{100}*$ ?15625

=?  $15625\left(1-\frac{20}{100}\right)$

=?  $15625*\left(1-\frac{1}{5}\right)$

=?  $15625*\frac{4}{5}$

Similarly,

Depreciated value after 2nd year =? 15625 $*{\left(\frac{4}{5}\right)}^2$ and so on.

This, Depreciated value at end of 5 years

=? 15625 $*{\left(\frac{4}{5}\right)}^5$

=? 15625 $*\frac{1024}{3125}$

=? 5120