Induction help needed! :P

Using the principle of mathematical induction, prove that $\frac{d^{n}a}{dx^{n}}=\frac{(-1)^{n}n!}{(x+5)^{n+1}}$ for $a=\frac{1}{x+5}$ .

Just explain me the inductive step! Thanks :)

#Calculus #MathematicalInduction #Induction

Easy Math Editor

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Comments

CASE I ( $n=1$ ): $\large \frac{d}{dx} \frac{1}{x+5} = \frac{-1}{(x+5)^2} = \frac{(-1)^{1} \cdot 1!}{(x+5)^{1+1}} \Rightarrow$ TRUE.

CASE II ( $n=k$ ): $\large \frac{d^k}{dx^k} \frac{1}{x+5} = \frac{(-1)^k \cdot k!}{(x+5)^{k+1}} \Rightarrow$ ASSUMED TRUE.

CASE III ( $n=k+1$ ): $\large \frac{d}{dx}[\frac{d^k}{dx^k} \frac{1}{x+5}] = \frac{d}{dx}[\frac{(-1)^k \cdot k!}{(x+5)^{k+1}}];$

or $\large \frac{d^{k+1}}{dx^{k+1}} \frac{1}{x+5}= \frac{(-1)^k \cdot -(k+1) \cdot k!}{(x+5)^{k+2}};$

or $\large \frac{d^{k+1}}{dx^{k+1}} \frac{1}{x+5} = \frac{(-1)^{k+1} \cdot (k+1)!}{(x+5)^{(k+1)+1}} \Rightarrow$ TRUE.

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$

tom engelsman - 3 weeks, 3 days ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$