Series Expansions and Coefficients

Consider given the series, $1+\left( 1+x \right)+{{\left( 1+x \right)}^{2}}+{{\left( 1+x \right)}^{3}}+...............{{\left( 1+x \right)}^{n}}$

Given series is G.P. which have first term $a=1$,comman ratio $r=1+x$ and number of term $=n$

So sum is,

${{s} _{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{[{{\left( 1+x \right)}^{n}}-1]}{1+x}={{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$

Now rth term of ${{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$ is $={}^{n-1}{{C} _{r}}{{x}^{r}}-{}^{-n}{{C} _{r}}.{{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right){{x}^{r}}$

Hence, coefficient of ${{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right)$

The Wallis formula is a formula for calculating the area of a circle. It was discovered by John Wallis in the 17th century, and is related to the work of Madhava of Sangamagrama on infinite series.

Madhava of Sangamagrama developed a series expansion for the sine function, which is given by $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$. This series converges for all (x).

The general formula for the Madhava Series expansion of the sine function is $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$

The Fourier series expansion of a function is an expansion of the function in terms of a series of sine and cosine functions.

Using the binomial theorem, we can expand ((1+x)^{10}) as (\sum_{k=0}^{10} \binom{10}{k} x^k). The coefficient of (x^3) corresponds to (k=3), so we have (\binom{10}{3} = \frac{10!}{3!7!} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1} = 120).

Using the binomial theorem, we can expand ((2x-3)^{8}) as (\sum_{k=0}^{8} \binom{8}{k} (2x)^{8-k}(-3)^k). The coefficient of (x^5) corresponds to (k=3), so we have (\binom{8}{3} (2)^{8-3}(-3)^3 = \frac{8!}{3!5!} \cdot 2^5 \cdot (-3)^3 = 2720).

Using the binomial theorem, we can expand ((x+2)^{12}) as (\sum_{k=0}^{12} \binom{12}{k} x^{12-k}2^k). The coefficient of (x^6) corresponds to (k=6), so we have (\binom{12}{6} 2^6 = \frac{12!}{6!6!} \cdot 2^6 = \frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1} \cdot 64 = 924).

Using the binomial theorem, we can expand ((3x^2-2)^{6}) as (\sum_{k=0}^{6} \binom{6}{k} (3x^2)^{6-k}(-2)^k). The coefficient of (x^4) corresponds to (k=2), so we have (\binom{6}{2} (3x^2)^{6-2}(-2)^2 = \frac{6!}{2!4!} \cdot 3^4x^8 \cdot 4 = 1296x^8).

The Taylor series expansion is a method for approximating functions using infinite series.

The coefficients of the expansion of (x + y)^n are given by the entries in the (n + 1)th row of Pascal's Triangle. Therefore, the coefficients of the expansion of (x + y)^6 are given by the entries in the 7th row of Pascal's Triangle.