\(\mathrm{e}^{-2 x}=1-\frac{2 x}{1 !}+\frac{(2 x)^{2}}{2 !}-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{4}}{4 !}-\frac{(2 x)^{5}}{5 !}+\ldots\)
\(\text { Coefficient of } x^{5} \text { is } \frac{-2^{5}}{5 !}=\frac{-32}{120}=\frac{-4}{15}\)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)

S₁₀ = 52, S₁₅ = 77, S₂₀ =?
\({S}_{10}={10\over 2}(2a(10-1)d)=52\)
= 5 (2a + 9d) = 52
\(2a+9d={52\over 5}\)
\({S}_{15}={15\over 2}(a+(15-1)d)=77\)
\(={15\over 2}(2a+14d)=77\)
\(2a+14d={154\over 15}\)
\(2a+9d={52\over 2}\)
(2)-(1) \(\Rightarrow\)    \({2a+14d={154\over 15}\over 5d={154\over 15}-{52\over 2}={154-156\over 15}}\)
\(5d={-2\over 15}\Rightarrow d={-2 \over 75}\)
substituting in (1) ​​​​​​​\(\Rightarrow\) \(2a-{18\over 75}={52\over 5}\)
\(2a={52\over 5}+{18\over 75}={780+18\over 75}\)
\(2a={798\over 75}\Rightarrow a={399\over 75}={133 \over 25}\)
\({S}_{20}={20 \over 2}\left[ \left( {798 \over 75} \right)+19\left( {-2\over 75} \right) \right]\)
\(=10{(798-38)\over 75}=10\left( {760\over 75} \right)\)
\(={1520\over 15}={304\over 3}\)

Given series is \(\sqrt { 3 } +\sqrt { 75 } +\sqrt { 243 } +.... .\) and \(S_n =435\sqrt { 3 }\)
Given series is \(1(\sqrt3)+5(\sqrt3)+9(\sqrt3)+...\)
Here a = √3, d = 4√3
∴ The given series an arithmetic progression
\(∴\ S_n={n\over2}[2a+(n-1)d]\)
\(435\sqrt3={n\over2}[2\sqrt3 +(n -1)4\sqrt3]\)  [∵ given S_n = 435√3J]
\(435\sqrt3={n\over2}[2\sqrt3+4n\sqrt3-4\sqrt4]\)

\(⇒\ 435\sqrt3={n\over2}[4n\sqrt3-2\sqrt3]\)
\(⇒\ 435\sqrt3=2{\sqrt3.n\over2}[2n-1]\)
⇒ 435 = 2n²-n
⇒ 2n²- n - 435 = 0
⇒ (n = 15)(2n + 29) = 0
⇒ \(n-15\ or\ n={-29\over2}\) which is not possible
⇒ n = 15

If we take y = 3x, then \({1\over (1+3x)^2}={1\over (1+y)^2}\)
Now \({1\over(1+y)^2}\) can be expanded using binomial theorem in powers of y. The expansion is valid only for values of y satisfying lyl < 1.
Replacing y by 3x we can get an expansion of \({1\over (1+3x)^2}.\) 
The expansion is valid only for values of x satisfying |3xl< 1; that is the expansion is valid only for values of x satisfying Ixl < \(1\over3\)
\({1\over (1+3x)^2}=(1+3x)^{-2}\)
\(=1-2(3x)+{2(2+1)\over 2!}(3x)^2-{2(2+1)(2+2)\over3!}(3x)^3+{2(2+1)(2+2)(2+3)\over 4!}(3x)^4-......\)
Hence, \({1\over (3+2x)^2}=1-6x+27x^2-108x^3+405x^4-...,|x|<{1\over 3}\)

\(\frac { 1 }{ 5+x } \) = (5 + x)^-1 
= \({ 5 }^{ -1 }{ \left( 1+\frac { x }{ 5 } \right) }^{ -1 }\)
= \(\frac { 1 }{ 5 } \left[ 1\left( -1 \right) \left( \frac { x }{ 5 } \right) +{ \left( \frac { x }{ 5 } \right) }^{ 2 }-{ \left( \frac { x }{ 5 } \right) }^{ 3 }+... \right] \)
= \(\frac { 1 }{ 5 } \left[ 1-\frac { x }{ 5 } +\frac { { x }^{ 2 } }{ 25 } -\frac { { x }^{ 3 } }{ 125 } +... \right] \)
The expansion is valid only if \(\left| \frac { x }{ 5 } \right| <1\ is\left| x \right| <5.\)

(10 -1)⁷  (a -b)ⁿ = nC₀ aⁿ b⁰ - nC₁ a^n-1 b¹ +... nC_n a⁰ bⁿ, n \(\in\) N
= 10⁷ - 7C₁ 10⁶ (1) + 7C₂ 10⁵ (1)² - 7C₃ 10⁴ (1)³ +7C₄ (10)³ (1)⁴ - 7C₅(10)² (1)⁵ + 7C₆(10)¹(1)⁶ - (1)⁷
= 10000000 - 7(1000000 ) + \(\frac { 7\times 6 }{ 2\times 1 } \)(100000) -\(\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } \) 10000 + \(\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } \) 1000
= - \(\frac { 7\times 6 }{ 2\times 1 } \) (100) + 7(10) - 1  
= 10000000 - 7000000 + 21.00000 -350000 + 35000 - 2100 + 70 - 1
= 4782969

As n = 7 which is odd, the terms containing x⁴y³ and x³y⁴ are the two middle terns.
They are ⁷C₃ x⁴y³ and ⁷C₄x³y⁴ which are equal 35x⁴y³ and 35x³y⁴.

\({ a }_{ n }=\begin{cases} n+1\quad if\quad n\quad is\quad odd \\ n\quad \quad if\quad n\quad is\quad even \end{cases}\)
a₁ = 1 + 1 = 2, a2 = 2, a₃ = 3 + 1 = 4
a₄ = 4, a₅ = 5 +1 = 6, a₆ = 6
hence the first 6 terms are 4, 2, 2, 4, 6, 6...

\({ a }_{ 1 }=\frac { \left( 1+1 \right) \left( 1+2 \right) }{ \left( 1+3 \right) (1+4) } =\frac { 2(3) }{ 4(5) } =\frac { 6 }{ 20 } =\frac { 3 }{ 10 } \)
\({ a }_{ 2 }=\frac { \left( 2+1 \right) \left( 2+2 \right) }{ \left( 2+3 \right) (2+4) } =\frac { 3(4) }{ 5(6) } =\frac { 12 }{ 30 } =\frac { 2 }{ 5 } \)
\({ a }_{ 3 }=\frac { \left( 3+1 \right) \left( 3+2 \right) }{ \left( 3+3 \right) (3+4) } =\frac { 4(5) }{ 6(7) } =\frac { 10 }{ 21 } \)
\({ a }_{ 4 }=\frac { 5(6) }{ 7(8) } =\frac { 15 }{ 28 } \)
\({ a }_{ 5 }=\frac { 6(7) }{ 8(9) } =\frac { 7 }{ 12 } \)
\({ a }_{ 6 }=\frac { 7(8) }{ 9(10) } =\frac { 28 }{ 45 } \)
The sequence is \(\frac { 3 }{ 10 } ,\frac { 2 }{ 5 } ,\frac { 10 }{ 21 } ,\frac { 15 }{ 28 } ,\frac { 7 }{ 12 } ,\frac { 28 }{ 45 } \)
None of A.P, G.P or H.P