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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 \( (1+x)^{n} \) . Let's call it equation \( I \) , where \( x \) is a complex number and \( n \) is a whole number .

\[\ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\dbinom{n}{2}x^{2}+\cdots+\dbinom{n}{n-1}x^{n-1}+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r} \]

Now let's integrate it once with respect to \(x\) .

\[\ \int_{0}^{x}(1+x)^{n} dx = \int_{0}^{x} \dbinom{n}{0}x^{0} dx +\int_{0}^{x} \dbinom{n}{1}x^{1} dx +\cdots+\int_{0}^{x} \dbinom{n}{n}x^{n}dx = \displaystyle\sum_{r=0}^{n}\int_{0}^{x} \dbinom{n}{r}x^{r}dx \]

\[ \dfrac{(1+x)^{n+1}-1}{n+1} = \dbinom{n}{0}\dfrac{x^{1}}{1}+\dbinom{n}{1}\dfrac{x^{2}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{n+1} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{x^{r+1}}{r+1} \]

Put \( x =1 \) and ....

\[ \boxed{ \dfrac{2^{n+1}-1}{n+1} = \dbinom{n}{0}\dfrac{1}{1}+\dbinom{n}{1}\dfrac{1}{2}+\cdots+\dbinom{n}{n}\dfrac{1}{n+1} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{r+1} }\]

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

\[S = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)\cdot(r+3)} \]

Let's start with \( I \)

\[\ \begin{align} (1+x)^{n} & = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\cdots+\dbinom{n}{n}x^{n} \quad \small\color{blue}{\text{0th integral}}\\ \int_{0}^{x}(1+x)^{n} dx & = \int_{0}^{x} \dbinom{n}{0}x^{0} dx +\int_{0}^{x} \dbinom{n}{1}x^{1} dx +\cdots+\int_{0}^{x} \dbinom{n}{n}x^{n}dx \\ \dfrac{(1+x)^{n+1}-1}{n+1} & = \dbinom{n}{0}\dfrac{x^{1}}{1}+\dbinom{n}{1}\dfrac{x^{2}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{n+1} \quad \small\color{blue}{\text{1st integral}} \\ \dfrac{x(1+x)^{n+1}-x}{n+1} & = \dbinom{n}{0}\dfrac{x^{2}}{1}+\dbinom{n}{1}\dfrac{x^{3}}{2}+\cdots+\dbinom{n}{n}\dfrac{x^{n+2}}{n+1} \quad \small\color{blue}{\text{times x}} \\ \int_{0}^{x} \dfrac{x(1+x)^{n+1}-x}{n+1}dx & = \int_{0}^{x} \dbinom{n}{0}\dfrac{x^{2}}{1}dx +\int_{0}^{x}\dbinom{n}{1}\dfrac{x^{3}}{2}dx+\cdots+\int_{0}^{x}\dbinom{n}{n}\dfrac{x^{n+2}}{n+1}dx \\ \dfrac{1}{n+1}\int_{0}^{x}x(1+x)^{n+1}dx -\dfrac{x^2}{2(n+1)}& = \dbinom{n}{0}\dfrac{x^3}{1 \cdot 3}+\dbinom{n}{1}\dfrac{x^4}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{x^{n+1}}{(n+1)\cdot(n+3)} \quad \small\color{blue}{\text{2nd integral}} \\ \dfrac{1}{n+1}\underbrace{\int_{0}^{1}x(1+x)^{n+1}dx}_{I_{1}} -\dfrac{1}{2(n+1)} & = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} \quad \small\color{blue}{\text{put x=1}} \end{align} \]


\[\ \begin{align} I_{x} & = \int_{0}^{x}x(1+x)^{n+1}dx \quad \quad \small\text{put } 1+x=t\\ & = \int_{1}^{x+1} (t-1)t^{n+1}dt \\ & = \int_{1}^{x+1} t^{n+2}-t^{n+1}dt \\ I_{x} & = \dfrac{(x+1)^{n+3}}{n+3}-\dfrac{(x+1)^{n+2}}{n+2}-\dfrac{1}{n+3}+\dfrac{1}{n+2}\\ I_{1} & = \dfrac{2^{n+3}-1}{n+3}-\dfrac{2^{n+2}-1}{n+2} \end{align} \]


\[\ \dfrac{1}{n+1}\left( \dfrac{2^{n+3}-1}{n+3}-\dfrac{2^{n+2}-1}{n+2} \right) -\dfrac{1}{2(n+1)} = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} \]

\[\ \boxed{ \dfrac{2^{n+3}(2n+1)-n^2-5n-4}{2(n+1)(n+2)(n+3)} = \dbinom{n}{0}\dfrac{1}{1 \cdot 3}+\dbinom{n}{1}\dfrac{1}{2\cdot 4}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+3)} } \]

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

\( 1) \hspace{2mm} \dbinom{n}{0}\dfrac{1}{1\cdot 2}+\dbinom{n}{1}\dfrac{1}{2\cdot 3}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)\cdot(n+2)} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)(r+2)} \)

\( 2) \hspace{2mm} \dbinom{n}{0}\dfrac{1}{1^2}+\dbinom{n}{1}\dfrac{1}{2^2}+\cdots+\dbinom{n}{n}\dfrac{1}{(n+1)^2} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}\dfrac{1}{(r+1)^2} \)

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
10 months ago

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Second problem's answer :

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

Harsh Shrivastava - 9 months, 4 weeks 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 - 9 months, 4 weeks ago

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