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, \({1\over (3+2x)^2}=1-6x+27x^2-108x^3+405x^4-...,|x|<{1\over 3}\)

Here a = 1 d = 3 and r \(={1\over5}\)
\(s_\infty={a\over 1-r}+{dr\over (1-r)^2}\)
\(={1\over 1-{1\over5}}+{3\times{1\over 5}\over({1-{1\over 5}})^2}\)
\(={5\over 4}+({3\over 5})({25\over 16})={35\over 16}\)

Let a, ar, ar², ... be the geometric series having the given properties.
Since the 4^th, 5^th and 6^th terms are ar³, ar⁴ and ar⁵, their product is a³r¹². Thus a³r¹² = 4096. 
Similarly a³r¹⁵ = 32768.
Therefore \({a^3r^15\over a^3{r}^{12}}={32768\over 4096}.\) 
Hence r³= 8. This implies that r = 2. a³r¹² = 4096 we have a³ = 1.
Therefore a = 1.
The sum of the first 8 terms is \({a(1-r^8)\over1-r}={1-2^8\over1-2}=255.\)

Since a = 4 and 4 + 8d = 7 we get \(d={3\over 8}.\) 
So the required 7 numbers are \(4{3 \over 8},4{6\over8},5{1\over8},5{4\over8},5{7\over8},6{2\over8},6{5\over8}.\)
Since a = 12 and \({ar}^{5}={3\over 8}\) we get \({1\over 32}\) and hence \(r={1\over 2}.\) 
Thus the required 4 numbers are \(6, 3, 1{1\over 2},{3\over 4}.\)

Let h_n be the harmonic progression and let \(a_n={1\over h_n}.\) 
Then a₅ = 19 and a₉ = 35.
As a_n's from an arithmetic progression, we have a + 4d = 19 and a + 8d = 35.
Solving these two equations, we get a = 3 and d = 4. 
Thus a₁₂ = a + 11d = 47.
Thus the 12^th term of the harmonic progression is \({1\over 47}.\)

\((x+y)^n:\)  The middle terms are \({{{T}_{n-1}}\over{2}}\) and \({{{T}_{n+1}}\over{2}}\)
Their coefficients are \(^n{C}_{{n+1\over2}}\) and \(^n{C}_{{n-1\over2}}\)
To prove they are equal   \(^n{C}_{{n+1\over2}}=^n{C}_{{n-\left( {n+1\over 2}\right)}}\) \([\because ^n{C}_{{r}}=^n{C}_{{n-r}}]\)
\(=^nC_{{2n-n+1\over2}}=^nC_{n-1\over2}\)

We have \({\left( 2x-{1\over 2x} \right)}^{4}\) = ⁴C₀(2x)⁴ \({\left(-{1\over 2x} \right)}^{0}\) +⁴C₁(2x)³\({\left(-{1\over 2x} \right)}^{1}\) + ⁴C₂(2x)²\({\left(-{1\over 2x} \right)}^{2}\)+⁴C₃(2x)¹\({\left(-{2\over x} \right)}^{3}\) + ⁴C₄(2x)⁰\({\left(-{1\over 2x} \right)}^{4}\)
= (2x)⁴ - 4(2x)³\({\left({1\over 2x} \right)}\) + 6(2x)²\({\left({1\over 2x} \right)}^{2}\)- 6(2x)\({\left({1\over 2x} \right)}^{2}+{\left({1\over 2x} \right)}^{4}\) 
\(=16x^4-16x^2+6-{3\over2x^2}+{1\over16x^4}\)

Let us take a = 3 and b = 2x in the binomial expansion of (a + b)¹⁰.
Then, x⁶ will appear in the term containing (2x)⁶ and nowhere else. So the term containing x⁶ is
\(^{10}{C}_{4}a^4b^6={10\times 9\times8\times 7\over4\times 3\times 2\times 1 }3^4{(2x)}^{6}=210\times3^4\times2^6x^6\)
So coefficient of x³ in the expansion of (3 + 2x)¹⁰ is 210 \(\times\) 3⁴2⁶

the terms in the numerator are 3, 5, 7 ..which forms an AP
t_n  = 3 + (n-1)2 = 3 +2n -2 = 2n + 1
the terms in the denominator are 1²2², 2²3², 3²4² 
t_n = [n(n+1)]²
∴  n^th terms of the given sequence is
= \(\frac { (2n+1) }{ [n(n+1)]^{ 2 } } =\frac { { n }^{ 2 }+2n+1-{ n }^{ 2 } }{ { n }^{ 2 }(n+1)^{ 2 } } \)
= \(\frac { ({ n }^{ 2 }+2n+1)-{ n }^{ 2 } }{ { n }^{ 2 }(n+1)^{ 2 } } =\frac { (n-1)^{ 2 }-{ n }^{ 2 } }{ { n }^{ 2 }(n+1)^{ 2 } } \)
= \(\frac { (n+1)^{ 2 } }{ { n }^{ 2 }(n+1)^{ 2 } } -\frac { { n }^{ 2 } }{ { n }^{ 2 }(n+1)^{ 2 } } =\frac { 1 }{ { n }^{ 2 } } -\frac { 1 }{ (n+1)^{ 2 } } \)
= \({ t }_{ n }=\frac { 1 }{ { n }^{ 2 }- } -\frac { 1 }{ (n+1)^{ 2 } } \)

Let a_n denote the n^thterm of the given series.
Then a_n \(={1\over n^2+5n+6}\) By using partial fraction, we get a_{n \(={1\over n+2}-{1\over n+3}\)}
Let S_n denote the sum of first n terms of the given series. Then
\(S_n=a_1+a_2+...+a_n=({1\over3}-{1\over 4})+({1\over4}-{1\over 5})+({1\over5}-{1\over 6})+....+({1\over n+2}-{1\over n+3})={1\over 3} -{1\over n+3}\)
But as n tends to infinity, \({1\over n+3}\) tends to zero and hence \({1\over 3}-{1\over n+3}\) tends to \({1\over 3}\). In other words S_n tends to \({1\over 3}\)
Thus\(\sum _{ n=1 }^{\infty }{1\over n^2+5n+6 } ={1\over 3}\)