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.

\(\require{color} \displaystyle \)

$$ (1 + x)^n = \binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n $$

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}

\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} \\ \)

Absolute Value Algebra Arithmetic Mean Arithmetic Sequence Binomial Expansion Binomial Theorem Chain Rule Circle Geometry Common Difference Common Ratio Compound Interest Cyclic Quadrilateral Differentiation Discriminant Double-Angle Formula Equation Exponent Exponential Function Factorials Functions Geometric Mean Geometric Sequence Geometric Series Inequality Integration Integration by Parts Kinematics Logarithm Logarithmic Functions Mathematical Induction Polynomial Probability Product Rule Proof Quadratic Quotient Rule Rational Functions Sequence Sketching Graphs Surds Transformation Trigonometric Functions Trigonometric Properties VCE Mathematics Volume