We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
167
2
avatar

Finished and Deleted

 Sep 18, 2018
edited by Guest  Sep 19, 2018
 #1
avatar+5226 
+1

I think this reads

 

\(x \geq -1,~\text{ Prove using induction that }(1+x)^n \geq 1+nx\)

 

\(P_1 = (1+x)^1 \geq 1+(1)x \\ \\ P_1 = (1+x) \geq (1+x) = TRUE\)

 

\(\text{Assume }P_n \text{ is TRUE} \\ \\ (1+x)^{n+1} = (1+x)^n(1+x) \geq (1+n x)(1+x) =\\ 1+(n+1)x+nx^2 \geq 1+(n+1)x \\ \\ \text{Thus }P_n \Rightarrow P_{n+1} \bigtriangleup\)

.
 Sep 18, 2018
 #2
avatar
0

THANKS SO MUCH!!!

 Sep 19, 2018

7 Online Users

avatar