Integrating Binomial Expansions

Integrating Binomial Expansion is being used for evaluating certain series or expansions by substituting particular values after integrating binomial expansion. It is important to find a suitable number to substitute for finding the integral constant if done in indefinite integral. If the definite integral is used, then it is important to set the upper and lower limits.
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$$ (1 + x)^n = \binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n $$

Question 1

Show that \( \displaystyle 1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + (-1)^n\frac{1}{n+1}\binom{n}{n} = \frac{1}{n + 1} \) using indefinite integral.

\( \displaystyle \begin{aligned}
\int \bigg[\binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n \bigg]dx &= \int (1 + x)^n dx \\
x + \frac{1}{2}\binom{n}{1}x^2 + \frac{1}{3}\binom{n}{2}x^3 + \cdots + \frac{1}{n+1}\binom{n}{n}x^{n+1} &= \frac{1}{n+1}(1 + x)^{n+1}+C \\
0 + 0 + 0 + \cdots + 0 &= \frac{1}{n+1} + C &\color{green}\text{substitute } x=0 \\
C &= -\frac{1}{n+1} \\
x + \frac{1}{2}\binom{n}{1}x^2 + \frac{1}{3}\binom{n}{2}x^3 + \cdots + \frac{1}{n + 1}\binom{n}{n}x^{n + 1} &= \frac{1}{n + 1}(1 + x)^{n + 1}-\frac{1}{n + 1} \\
-1 + \frac{1}{2}\binom{n}{1} – \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= -\frac{1}{n + 1} &\color{green}\text{substitute } x=-1 \\
1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots – \frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= \frac{1}{n + 1} &\color{green}\text{change signs} \\
1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots – \frac{1}{n+1}\binom{n}{n}(-1)^n(-1) &= \frac{1}{n + 1} \\
\therefore 1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^n &= \frac{1}{n + 1} \\
\end{aligned} \)

Question 2

Show that \( \displaystyle 1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + (-1)^n\frac{1}{n+1}\binom{n}{n} = \frac{1}{n + 1} \) using definite integral.

\( \displaystyle \begin{aligned}
\int_{0}^{-1} \bigg[\binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n \bigg]dx &= \int_{0}^{-1} (1 + x)^n dx \\
\bigg[x + \frac{1}{2}\binom{n}{1}x^2 + \frac{1}{3}\binom{n}{2}x^3 + \cdots + \frac{1}{n+1}\binom{n}{n}x^{n+1}\bigg]_{0}^{-1} &= \bigg[\frac{1}{n+1}(1 + x)^{n+1}\bigg]_{0}^{-1} \\
-1 + \frac{1}{2}\binom{n}{1} – \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= -\frac{1}{n + 1} \\
1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots – \frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= \frac{1}{n + 1} &\color{green}\text{change signs} \\
1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots – \frac{1}{n+1}\binom{n}{n}(-1)^n(-1) &= \frac{1}{n + 1} \\
\therefore 1 – \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^n &= \frac{1}{n + 1} \\
\end{aligned} \\ \)

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