In (1+x)ⁿ, n = n, x = 1, A = x
general terms t_r+1 = nC_r, X^n-r a^r
t_r+1 = nC_r, (1)^n-r a^r
t_r+1 = nC_r x^r
∴ Co-efficient of X^r is nC_r
Putting r = n -r in (1) we get
T_n-r+1 = nC_n-r X^n-r
Co-efficient of x^n-r is nC_n-r
But nC_r = nC_n-r Using the property of combination
ஃ Coefficients of x^r and Co-efficients of x^n-r  are equal

we have \({ e }^{ x }=1+\frac { x }{ 1! } +\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 4 } }{ 4! } +..\)
\({ e }^{ \frac { 1 }{ 2 } x }=1+\frac { \left( \frac { 1 }{ 2 } x \right) }{ 1! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 2 } }{ 2! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 3 } }{ 3! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 4 } }{ 4! } +...\)
\(=1+\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 8 } +\frac { { x }^{ 3 } }{ 48 } +\frac { { x }^{ 4 } }{ 388 } +\frac { { x }^{ 5 } }{ 32\times 5 } +...\)
\({ e }^{ \frac { 1 }{ 2 } x }=1+\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 8 } +\frac { { x }^{ 3 } }{ 48 } +\frac { { x }^{ 4 } }{ 388 } +\frac { { x }^{ 5 } }{ 3840 } +...\)

we have \({ e }^{ -x }=1+\frac { x }{ 1! } +\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 4 } }{ 4! } +\frac { { x }^{ 5 } }{ 5! }+..\)
\(\therefore { e }^{ -2x }=1-\frac { \left( 2x \right) }{ 1! } +\frac { { \left( 2x \right) }^{ 2 } }{ 2! } -\frac { { \left( 2x \right) }^{ 3 } }{ 3! } +\frac { { \left( 2x \right) }^{ 4 } }{ 4! } -\frac { { \left( 2x \right) }^{ 5 } }{ 5! } +.....\)
\(=1-2x+\frac { { 4x }^{ 2 } }{ 2 } -\frac { { 8x }^{ 3 } }{ 6 } +\frac { 16x^{ 4 } }{ 24 } -\frac { 32x^{ 5 } }{ 120 } +......\)
\({ e }^{ -2x }=1-2x+{ 2x }^{ 2 }-\frac { { 4x }^{ 3 } }{ 3 } +\frac { 2 }{ 3 } { x }^{ 4 }-\frac { 4 }{ 15 } { x }^{ 5 }+......\)

We have \({ e }^{ 5x }=1+\frac { x }{ 1! } +\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 4 } }{ 4! } +..\)
\({ e }^{ 5x }=1+\frac { 5x }{ 1! } +\frac { { \left( 5x \right) }^{ 2 } }{ 2! } +\frac { { \left( 5x \right) }^{ 3 } }{ 3! } +\frac { { \left( 5x \right) }^{ 4 } }{ 4! } +\frac { { \left( 5x \right) }^{ 5 } }{ 5! } +.....\)
 \(=1+5x+\frac { { 25x }^{ 2 } }{ 2 } +\frac { { 125x }^{ 3 } }{ 6 } +\frac { 6{ 25x }^{ 4 } }{ 24 } +\frac { 6{ 25x }^{ 5 } }{ 24 } +....\)

\((x+2)^{-2\over3}=(2+x)^{-2\over3}\)
\(=2^{-2\over3}\left(1+{x\over2}\right)\left[∵(1+x)^{-p\over q}=1-\left(p\over q\right)x+{\left(p\over q\right)\left({p\over q}-1\right)\over2!}x^2-{\left(p\over q\right)\left({p\over q}-1\right)\left({p\over q}-2\right)\over3!}x^3+... \right]\)
\(={1\over 2^{2\over3}} \left( 1-{2\over3}\left(x\over2\right)+{\left(-{2\over3} \right)\left(-{2\over3}-1 \right)\over2!} \left( x\over2 \right)^2+{\left(-{2\over3} \right)\left(-{2\over3}-1\right)\left(-{2\over3}-2 \right)\over3!\left({x\over2} \right)^3}+{\left(-{2\over3} \right)\left(-{2\over3} -1 \right)\left(-{2\over3}-2 \right)\left(-{2\over3}-3 \right)\over 4!}\left({x\over2} \right)^4+{\left( - {2\over3} \right)\left( - {2\over3}-1 \right)\left( -{2\over3}-2 \right)\left(- {2\over3}-3 \right)\left( - {2\over3} -4 \right)\over5!}\left( {x\over2} \right)^5+.... \right)\)\(={1\over 2^{2\over3}}\left[ 1-{x\over3}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\over2 }{x^2\over4}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\left( -{8\over3}\right)\over6}{x^3\over8}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\left( -{8\over3}\right)\left( -{11\over3}\right)\over24}{x^4\over16}+... \right]\)
\(={ 2 }^{ \frac { -2 }{ 3 } }\left[ 1-\frac { x }{ 3 } +\frac { { 5x }^{ 2 } }{ 36 } -\frac { 5 }{ 81 } .{ x }^{ 3 }+\frac { 55 }{ 1944 } { x }^{ 4 }-.. \right] \)

Let T_n be the nth term of the given sequence.
Given sequence is \({1\over1},{4\over3},{7\over9},{10\over27}....\)
Consider the terms in the numerator
1, 4, 7, 10,...
Here a = 1, d = 3
The terms in the denominator are  \({1\over3^0},{1\over3^1},{1\over 3^2}\), which is a G.P with  \(r={1\over3}\)
∴ The given sequence can be written in the form of a, (a + d)r, (a + 2d)r²,(a + 3d), r³, ...
This is an arithmetic - geometric progression.
∴ T_n = [a+(n-1)d]r^n-1
\(=[1+(n -1)3]\left(1\over3\right)^{n-1}\)
\(=({1+3n-3})\left(1\over 3^{n-1}\right)={3n-2\over 3^{n-1}}\)
\(∴\ T_n={3n-2\over 3^{n-1}}\)
Let S_n be the sum to n terms of the given sequence
\(S_n=\sum_{k=1}^n{3k-2\over 3^{k-1}}\)
\(={\sum_{k=1}^n3k-2.{1\over{\sum_{k=1}^n}3^{k-1}}}\)
\(= 3[1+ 2 + 3+ ...+ n] - 2n \left[ 1\over3^0+3^2+...+3^{r-1}\right]\)
\(=\left[ 3{n(n+1)\over2}-2n\right]\left[ 1\over 1\left(3^n-1\over 3-1\right)\right]\)
 \(=\left[{3n^2+3n\over2}-2n\right]\left[2\over 3^n-1\right]={3n^2+3n-4n\over2}\times{2\over3^n-1}\)
\(\frac { { 3n }^{ 2 }-n }{ { 3 }^{ n }-1 } =\frac { n\left( n-1 \right) }{ { 3 }^{ n }-1 } \)

=[(x-a)ⁿ = xⁿ + ⁿC₁x^n-1(-a)¹+ⁿC₂x^n-1(-a)²+.....(-a)ⁿ]
= \(​​​​​​​​\left({ 2x }^{ 2 } \right) ^{ 3 }+3C_{ 1 }\left( { 2x }^{ 2 } \right) ^{ 2 }\left( \frac { 3 }{ x } \right) ^{ 1 }+{ 3C }_{ 2 }\left( { 2x }^{ 2 } \right) ^{ 1 }\left( \frac { 3 }{ x } \right) ^{ 2 }+\left( -\frac { 3 }{ x } \right) ^{ 3 }\)
= \({ 8x }^{ 6 }+3\left( { 4x }^{ 4 } \right) \left( -\frac { 3 }{ x } \right) +\frac { 3\times 2 }{ 2\times 1 } \left( { 2x }^{ 2 } \right) \left( \frac { 9 }{ { x }^{ 2 } } \right) -\frac { 27 }{ { x }^{ 3 } } \)
= \({ 8x }^{ 6 }-{ 36x }^{ 3 }+54-\frac { 27 }{ { x }^{ 3 } } \)

\(log\left(1+3\over1-3x\right)=log(1+3x)-log(1-3x)\)[using logarithmic, quotient rule]
\(=\left[ 3x-{(3x)^2\over2}+{(3x)^3\over3}-{(3x)^4\over4}+...\right]-\left[-3x-{(3x)^2\over2}-{(3x)^3\over3}-{(3x)^4\over4}-...\right]\)
\(=\left[ 3x-{9x^2\over2}+{27x^3\over3}-{81x^4\over4}+...\right]-\left[ -3x-{9x^2\over2}-{27x^3\over3}-{81x^4\over4}+...\right]\)
\(=3x-{9x^2\over2}+{27x^3\over3}-{81x^4\over4}x^4+...+3x+{9x^2\over2}+{27x^3\over3}+{81x^4\over4}+...\)
\(=2\left[ 3x+{37x^3\over3}+{243x^5\over5}+{2187x^7\over7}+...\right]\)
This series is valid only when \(\left| 3x \right| <1\Rightarrow \left| x \right| <\frac { 1 }{ 3 } \)
Hence, this series is valid in the interval \(-\frac { 1 }{ 3 }< x<{1\over3}\)

According to the given information, we have the following diagram.

Distance travelled to bring first ball = 24 + 24 = 2 \(\times\) 24 = 48 m
Distance travelled to bring second ball = 2 (24 + 4) = 2(28) = 56 m
Distance travelled to bring third ball = 2 (24 + 4 + 4) = 2(32) = 64 m
\(\therefore\) The series of distances are 48, 56, 64 ...
Here a = 48, d = 56 - 48 = 8 and n = 20.
To find the total distance that he run in bringing back all balls, we have to find the sum of 20 terms of the above series
\(\therefore\) \({ S }_{ 20 }=\frac { 20 }{ 2 } \left[ 2\left( 48 \right) +19\left( 8 \right) \right] \)
= 10[96 + 152]
= 10[248]
S₂₀ = 2480 m.

Suppose the loan in cleared in n months. Clearly the amount forms an. A.P. with a = 20 and d = 15
∴ Sum of the amounts = 3250
S_n = 3250

\(⇒\ {n\over2}[2a + (n -1)d]=3250\)
\(⇒\ {n\over2}[40+(n-1)15]=3250\)
⇒ n(40 + 15n - 15) = 6500
⇒ n (15n + 25) 6500
⇒ 15n² + 25n = 6500
⇒ 15n² + 25n = 6500
⇒ 3n² + 5n - 1300 = 0
⇒ (n - 20) (3n + 65) = 0
⇒ n = 20 or \(n={-65\over 3}\) which is not possible
∴ n = 20
Thus, the amount is cleared in 20 months.