Laws of Binomial theorem (2.2)

Hello everyone , this is part 2.2 of binomial theorem ,a continuation of 2.1 where we used differentiation to get desired results . In this part we will use Integration , a complicated tool in mathematics which helps in calculating areas and averaging continuous functions .Only basic understanding of integrals and U-substitution is required to understand this note. By the way, check out wiki pages of brilliant on Definite integrals and Integration of Algebraic Functions , they are just awesome . Let's start with our cute little equation , expansion of (1+x)n (1+x)^{n} . Let's call it equation I I , where x x is a complex number and n n is a whole number .

 (1+x)n=(n0)x0+(n1)x1+(n2)x2++(nn1)xn1+(nn)xn=r=0n(nr)xr\ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\dbinom{n}{2}x^{2}+\cdots+\dbinom{n}{n-1}x^{n-1}+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r}

Now let's integrate it once with respect to xx .

 0x(1+x)ndx=0x(n0)x0dx+0x(n1)x1dx++0x(nn)xndx=r=0n0x(nr)xrdx\ \int_{0}^{x}(1+x)^{n} dx = \int_{0}^{x} \dbinom{n}{0}x^{0} dx +\int_{0}^{x} \dbinom{n}{1}x^{1} dx +\cdots+\int_{0}^{x} \dbinom{n}{n}x^{n}dx = \displaystyle\sum_{r=0}^{n}\int_{0}^{x} \dbinom{n}{r}x^{r}dx

(1+x)n+11n+1=(n0)x11+(n1)x22++(nn)xn+1n+1=r=0n(nr)xr+1r+1 \dfrac{(1+x)^{n+1}-1}{n+1} = \dbinom{n}{0}\dfrac{x^{1}}{1}+\dbinom{n}{1}\dfrac{x^{2}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{n+1} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{x^{r+1}}{r+1}

Put x=1 x =1 and ....

2n+11n+1=(n0)11+(n1)12++(nn)1n+1=r=0n(nr)1r+1 \boxed{ \dfrac{2^{n+1}-1}{n+1} = \dbinom{n}{0}\dfrac{1}{1}+\dbinom{n}{1}\dfrac{1}{2}+\cdots+\dbinom{n}{n}\dfrac{1}{n+1} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{r+1} }

Similarly we can get loads of new series . Let's try to get

S=(n0)113+(n1)124++(nn)1(n+1)(n+3)=r=0n(nr)1(r+1)(r+3)S = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)\cdot(r+3)}

Let's start with I I

 (1+x)n=(n0)x0+(n1)x1++(nn)xn0th integral0x(1+x)ndx=0x(n0)x0dx+0x(n1)x1dx++0x(nn)xndx(1+x)n+11n+1=(n0)x11+(n1)x22++(nn)xn+1n+11st integralx(1+x)n+1xn+1=(n0)x21+(n1)x32++(nn)xn+2n+1times x0xx(1+x)n+1xn+1dx=0x(n0)x21dx+0x(n1)x32dx++0x(nn)xn+2n+1dx1n+10xx(1+x)n+1dxx22(n+1)=(n0)x313+(n1)x424++(nn)xn+1(n+1)(n+3)2nd integral1n+101x(1+x)n+1dxI112(n+1)=(n0)113+(n1)124++(nn)1(n+1)(n+3)put x=1\ \begin{aligned} (1+x)^{n} & = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\cdots+\dbinom{n}{n}x^{n} \quad \small\color{#3D99F6}{\text{0th integral}}\\ \int_{0}^{x}(1+x)^{n} dx & = \int_{0}^{x} \dbinom{n}{0}x^{0} dx +\int_{0}^{x} \dbinom{n}{1}x^{1} dx +\cdots+\int_{0}^{x} \dbinom{n}{n}x^{n}dx \\ \dfrac{(1+x)^{n+1}-1}{n+1} & = \dbinom{n}{0}\dfrac{x^{1}}{1}+\dbinom{n}{1}\dfrac{x^{2}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{n+1} \quad \small\color{#3D99F6}{\text{1st integral}} \\ \dfrac{x(1+x)^{n+1}-x}{n+1} & = \dbinom{n}{0}\dfrac{x^{2}}{1}+\dbinom{n}{1}\dfrac{x^{3}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+2}}{n+1} \quad \small\color{#3D99F6}{\text{times x}} \\ \int_{0}^{x} \dfrac{x(1+x)^{n+1}-x}{n+1}dx & = \int_{0}^{x} \dbinom{n}{0}\dfrac{x^{2}}{1}dx +\int_{0}^{x}\dbinom{n}{1}\dfrac{x^{3}}{2}dx+\cdots+\int_{0}^{x}\dbinom{n}{n}\dfrac{x^{n+2}}{n+1}dx \\ \dfrac{1}{n+1}\int_{0}^{x}x(1+x)^{n+1}dx -\dfrac{x^2}{2(n+1)}& = \dbinom{n}{0}\dfrac{x^3}{1 \cdot 3}+\dbinom{n}{1}\dfrac{x^4}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{(n+1)\cdot(n+3)} \quad \small\color{#3D99F6}{\text{2nd integral}} \\ \dfrac{1}{n+1}\underbrace{\int_{0}^{1}x(1+x)^{n+1}dx}_{I_{1}} -\dfrac{1}{2(n+1)} & = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} \quad \small\color{#3D99F6}{\text{put x=1}} \end{aligned}

 Ix=0xx(1+x)n+1dxput 1+x=t=1x+1(t1)tn+1dt=1x+1tn+2tn+1dtIx=(x+1)n+3n+3(x+1)n+2n+21n+3+1n+2I1=2n+31n+32n+21n+2\ \begin{aligned} I_{x} & = \int_{0}^{x}x(1+x)^{n+1}dx \quad \quad \small\text{put } 1+x=t\\ & = \int_{1}^{x+1} (t-1)t^{n+1}dt \\ & = \int_{1}^{x+1} t^{n+2}-t^{n+1}dt \\ I_{x} & = \dfrac{(x+1)^{n+3}}{n+3}-\dfrac{(x+1)^{n+2}}{n+2}-\dfrac{1}{n+3}+\dfrac{1}{n+2}\\ I_{1} & = \dfrac{2^{n+3}-1}{n+3}-\dfrac{2^{n+2}-1}{n+2} \end{aligned}

 1n+1(2n+31n+32n+21n+2)12(n+1)=(n0)113+(n1)124++(nn)1(n+1)(n+3)\ \dfrac{1}{n+1}\left( \dfrac{2^{n+3}-1}{n+3}-\dfrac{2^{n+2}-1}{n+2} \right) -\dfrac{1}{2(n+1)} = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)}

 2n+3(2n+1)n25n42(n+1)(n+2)(n+3)=(n0)113+(n1)124++(nn)1(n+1)(n+3)\ \boxed{ \dfrac{2^{n+3}(2n+1)-n^2-5n-4}{2(n+1)(n+2)(n+3)} = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} }

For Practice try to find the sum of the following series:

1)(n0)112+(n1)123++(nn)1(n+1)(n+2)=r=0n(nr)1(r+1)(r+2) 1) \hspace{2mm} \dbinom{n}{0}\dfrac{1}{1\cdot 2}+\dbinom{n}{1}\dfrac{1}{2\cdot 3}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+2)} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)(r+2)}

2)(n0)112+(n1)122++(nn)1(n+1)2=r=0n(nr)1(r+1)2 2) \hspace{2mm} \dbinom{n}{0}\dfrac{1}{1^2}+\dbinom{n}{1}\dfrac{1}{2^2}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)^2} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)^2}

I Hope you enjoyed this note and If someone knows the answer to 2nd problem please post it in the comments section below. thank you :) . And Merry Christmas everyone !!!

Note by Sabhrant Sachan
4 years, 6 months ago

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Second problem's answer :

The problem burns down to 01(1+x)n+11x(n+1)dx\displaystyle \int_{0}^{1} \dfrac{(1+x)^{n+1} -1}{x(n+1)} dx

Harsh Shrivastava - 4 years, 6 months ago

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Exactly I went a little ahead and got this to be s=1/(n+1)*(1+sum (2^r-1^r)/r)where r varies from 2 to n+1.

Spandan Senapati - 4 years, 6 months ago

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