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

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 Algebraic Fractions Arithmetic Sequence Binomial Expansion Chain Rule Circle Geometry Common Difference Common Ratio Compound Angle Formula Compound Interest Cyclic Quadrilateral Differentiation Discriminant Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Inequality Integration Kinematics Logarithm Logarithmic Functions Mathematical Induction Perfect Square Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Quotient Rule Rational Functions Sequence Sketching Graphs Surds Transformation Trigonometric Functions Trigonometric Properties Volume

 



Your email address will not be published. Required fields are marked *