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

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2 \sqrt{2}}, \ldots \ldots \text { form } a \text { HP. }\)

Fourth root of 623 = \(\left( \frac { -2 }{ 625 } \right) \)
\(\left( \frac { -2 }{ 625 } \right) \)
\(={ \left[ 625\left( 1-\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }=5{ \left[ 1+\left( -\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }\)
\(={ 5\left[ 1+\frac { 1 }{ 4 } \left( \frac { -2 }{ 625 } \right) +\frac { \frac { 1 }{ 4 } \left( -\frac { 3 }{ 4 } \right) }{ 1.2 } { \left( \frac { -2 }{ 625 } \right) }^{ 2 } \right] }\)
Other terms will have more than seven zeroes after the decimal]
= 5[1-0.0008 - 0.0000009]
\(\sqrt [ 4 ]{ 623 } =4.9959955\)

Given series is \(1+2\left( {1 \over 5} \right)+3{\left( {1\over 5} \right)}^{2}+4{\left({1\over 5} \right)}^{2}+....\)   ...(1)
This is an arithmetico - geometric series with corresponding A.P. 1, 2, 3, 4, .... and G.P: \(1,{1 \over 5},\left( {1\over 5} \right)^2,{\left( {1\over 5} \right)}^{3},......\)
Let \(S=1+2\left({1\over 5} \right)+3{\left( {1\over 5} \right)}^{2}+4{\left({1\over 5} \right)}^{3}+....\)  ...(2)
Multiplying both sides by the common ratio \(={1\over 5}\) of G.P. we get,
\({1\over 5}S={1\over5}+2{\left( {1 \over 5} \right)}^{2}+3{\left( {1\over 5} \right)}^{3}+4{\left( {1\over 5} \right)}^{4}+....\)
(2) - (1), we get,
\(S-{1\over5}S=1+(2-1)\left( {1\over5} \right)(3-2){\left( {1\over 5} \right)}^{2}+(4-3){\left({1\over 5} \right)}^{2}+...\)
\(\Rightarrow\)     \(S-{1\over 5}S=1+\left[ {1\over5}+{\left( {1\over 5} \right)}^{2}+{\left( {1\over 5} \right)}^{3}+... \right]\)
\(=1+\frac { \frac { 1 }{5 } }{ 1-\frac { 1 }{ 5 } } \)    \(\left[ \because {S}_{\infty}={a \over 1-r} \right]\)
\(=1+\frac { \frac { 1 }{ 5 } }{ \frac { 4 }{ 5 } } =1+{1\over 4}={5\over 4}\)   \(\left[ \because |{1\over 5}| ={1\over 5}<1\right]\)
\(\Rightarrow\)   \(S={5\over 4}\times{5\over 5}={25\over16}\)

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

Let T_n be the nth term of the given series
Then T_n = 1 + 4 + 4² + 4³ + ...
\(=1\left(4^n-1\over 4-1\right)\)
\(={4^n-1\over 3}\)
Let S_n be the sum to n terms of the given series
Then \(S_n={\sum_{k=1}^n}T_k=\sum_{k=1}^n{4^n-3\over3}\)
\(⇒\ S_n={1\over3}\left[ \sum_{k=1}^n4^n-\sum_{k=1}^n3\right]\)
\(⇒\ S_n= {1\over3}[4^1+4^]+...+4^n-3^n\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]={1\over 3}\left[4{(4^n-1)-9n\over3}\right]\)

\({ S }_{ n }=\frac { 4 }{ 9 } \left[ \left( { 4 }^{ n }-1 \right) -n/3 \right] \)

(1 + x) + (1 + x + x²) + (1 + x + x² + x³) + ... up to n terms
= \(\frac{1-x^{2}}{1-x}+\frac{1-x^{3}}{1-x}+..\frac{1-x^{4}}{1-x}+...\) to n terms
= \(\frac{1}{1-x}[(1+1+1+... to\ n\ terms)-(x^{2}+x^{3}+x^{4}.... to\ n\ terms)\)
= \(\frac{1}{1-x}[n-\frac{x^{2}(1-x^{n})}{1-x}]\)

\(\frac{1}{2}\)

Let the number be a and b
∴ a + b = \(\frac { 13 }{ 6 } \)
Let A₁, A₂,..A_2n be the 2n A,M s between a and b ....(1)
= \(2n\left( \frac { a+b }{ 2 } \right) =n(a+b)=n\times \frac { 13 }{ 6 } (2)\)   using (1)
Also  A₁+ A₂+ A_2n = 2n+1(given)   ...(3)
From (2)  and (3),\(\frac { 13n }{ 6 } \) = 2n+1
⇒   13n = 12n + 6
⇒    n = 6
∴  No of A.M's inserted = 2n - 2(6) = 12

We have log (1 - x) = \(-x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 4 } }{ 4 } ....\)
\(\therefore \log { \left( 1-2x \right) } =-\left( 2x \right) -\frac { { \left( 2x \right) }^{ 2 } }{ 2 } -\frac { { \left( 2x \right) }^{ 3 } }{ 3 } -\frac { { \left( 2x \right) }^{ 4 } }{ 4 } +\frac { { \left( 2x \right) }^{ 5 } }{ 5 } -\frac { { \left( 2x \right) }^{ 6 } }{ 6 } +....\)
\(\log { \left( 1-2x \right) } =-2x-\frac { { 4x }^{ 2 } }{ 2 } -\frac { { 8x }^{ 3 } }{ 3 } -\frac { { 16x }^{ 4 } }{ 4 } -\frac { { 32x }^{ 5 } }{ 5 } -\frac { 6{ 4x }^{ 6 } }{ 6 } +....\)
This series is valid only when \(\left| 2x \right| <1\Rightarrow \left| x \right| <\frac { 1 }{ 2 } \)
Hence, this series is valid only in the interval \(-\frac { 1 }{ 2 }

(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

The general term in the expansion of (1 + x)^m is T_r+1 = ⁿC_r(1)^m-r x^r
On putting r = 2, we get T₃ = ^mC₂(1)^m-2 x² =  ^mC₂ x²
∴ Coefficient of x² = ^mC₂
Also, coefficient of x² in the expansion of (1+x)^m is 6
∴ ^mC₂ = 6 ⇒ \(\frac{m(m-1)}{2.1}=6 \Rightarrow m(m-1)=12\)
⇒ m(m-1) = 4.3
⇒ m = 4 

Since H is the H. M. between a and b,
\(we\quad get\quad H=\frac { 2ab }{ a+b } \quad ........(1)\)
\(LHS=(H-2a)(H-2b)\)
\(=\left( \frac { 2ab }{ a+b } -2a \right) \left( \frac { 2ab }{ a+b } -2b \right) \)
\(=\left( \frac { 2ab-2{ a }^{ 2 }-2ab }{ a+b } \right) \left( \frac { 2ab--2ab-{ ab }^{ 2 } }{ a+b } \right) \)
\(=\left( \frac { { -2a }^{ 2 } }{ a+b } \right) \left( \frac { { -2b }^{ 2 } }{ a+b } \right) =\left( \frac { { 4a }^{ 2 }{ b }^{ 2 } }{ { \left( a+b \right) }^{ 2 } } \right) { \left( \frac { 2ab }{ a+b } \right) }^{ 2 }\)
\(={ H }^{ 2 }[using\quad 1]\)

Given x = a + \(\frac{a}{r}+\frac{a}{r^{2}}+...\infty\)
\(=a(1+\frac{1}{r}+\frac{1}{r^{2}}+...+\infty)\)   \([\because S=\frac{a}{1-r}]\)
= \(a(\frac{1}{1-\frac{1}{r}})=a(\frac{r}{r-1})\)
x = \(\frac{ar}{r-1}\) --- (1)
y = \(b-\frac{b}{r}+\frac{b}{r^{2}}...b(\frac{1}{1-(\frac{-1}{r})})=\frac{b}{1+\frac{1}{r}}\)
y = \(\frac{br}{r+1}\)   --- (2)
z = c + \(\frac{c}{r^{2}}+\frac{c}{r^{4}}+..\) = \(\frac{c}{1-\frac{1}{r^{2}}}=\frac{cr^{2}}{r^{2}-1}\)  ---- (3)
ஃ \(\frac{xy}{z}=(\frac{ar}{r-1}.\frac{br}{r+1})/\frac{cr^{2}}{r^{2}-1}=\frac{abr^{2}}{r^{2}-1}\times\frac{r^{2}-1}{cr^{2}}=\frac{ab}{c}\)  [∵ using (1),(2) and (3)]
⇒ \(\frac{xy}{z}=\frac{ab}{c}\)

2018
Let a_n = 2018
then the first 6 terms are 2018, 2018, 2018, 2018, 2018, 2018
It is not an AP, GP, AGP and HP.

Let a_n = \(\frac { 2n+3 }{ 3n+4 } \)
\({ a }_{ 1 }=\frac { 2+3 }{ 3+4 } =\frac { 5 }{ 9 } \)
\({ a }_{ 2 }=\frac { 4+3 }{ 6+4 } =\frac { 7 }{ 10 } \)
\({ a }_{ 3 }=\frac { 6+3 }{ 9+4 } =\frac { 9 }{ 13 } \)
\({ a }4=\frac { 8+3 }{ 12+4 } =\frac { 11 }{ 16 } \)
\({ a }_{ 5 }=\frac { 10+3 }{ 15+4 } =\frac { 13 }{ 19 } \)
\({ a }_{ 6 }=\frac { 12+3 }{ 18+4 } =\frac { 15 }{ 22 } \)
\(\frac { 5 }{ 9 } ,\frac { 7 }{ 10 } ,\frac { 9 }{ 13 } ,\frac { 11 }{ 16 } ,\frac { 13 }{ 19 } ,\frac { 15 }{ 22 } ...\)
this is neither A.P, G.P nor AGP