Integrating Binomial Expansions
Integrating Binomial Expansion is used to evaluate 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 an indefinite integral. If the definite integral is used, it is important to set the upper and lower limits.
\(\require{color} \displaystyle \)
$$ (1 + x)^n = \binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n $$
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.
\( \displaystyle \begin{aligned} \require{AMSsymbols}
\int \bigg[\binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n \bigg]dx &= \int (1 + x)^n dx \\
x + \frac{1}{2}\binom{n}{1}x^2 + \frac{1}{3}\binom{n}{2}x^3 + \cdots + \frac{1}{n+1}\binom{n}{n}x^{n+1} &= \frac{1}{n+1}(1 + x)^{n+1}+C \\
0 + 0 + 0 + \cdots + 0 &= \frac{1}{n+1} + C &\color{green}\text{substitute } x=0 \\
C &= -\frac{1}{n+1} \\
x + \frac{1}{2}\binom{n}{1}x^2 + \frac{1}{3}\binom{n}{2}x^3 + \cdots + \frac{1}{n + 1}\binom{n}{n}x^{n + 1} &= \frac{1}{n + 1}(1 + x)^{n + 1}-\frac{1}{n + 1} \\
-1 + \frac{1}{2}\binom{n}{1}-\frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= -\frac{1}{n + 1} &\color{green}\text{substitute } x=-1 \\
1-\frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots-\frac{1}{n+1}\binom{n}{n}(-1)^{n+1} &= \frac{1}{n + 1} &\color{green}\text{change signs} \\
1-\frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots-\frac{1}{n+1}\binom{n}{n}(-1)^n(-1) &= \frac{1}{n + 1} \\
\therefore 1-\frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}(-1)^n &= \frac{1}{n + 1}
\end{aligned} \)
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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