0

I need to prove by induction that for every $n \ge 1$, $n\in\mathbb{N}$ and $a>0$ the following statement holds $$(1+a)^n \ge 1 + na + \dfrac{n(n-1)}{2}a^2$$

The statement is true for $n=1$: $$1+a \ge 1+a$$

Assume for some $n=k$: $$(1+a)^k \ge 1 + ka + \dfrac{k(k-1)}{2}a^2$$

Prove for $n=k+1$: $$(1+a)(1+a)^k \ge 1+(k+1)a+\dfrac{k(k+1)}{2}a^2$$

And I'm pretty much stuck here...

McLovin
  • 550

2 Answers2

2

The statement can be proven directly from Bernoulli's inequality but since you must use induction..

HINT

As Dr.Sonhard Graubner mentioned you should multiply by $a+1$ to obtain $$(1+a)^{k+1}\geq 1+a+ka(1+a)+\frac{k(k-1)(1+a)}{2}a^2$$

and now compare the RHS with $$1+(k+1)a+\dfrac{k(k+1)}{2}a^2$$

After a little algebra you will end up with an expression $ka^3(k-1)\ge0$ which holds true for $a\gt0,k\gt1$

MathematicianByMistake
  • 5,495
  • I don't get it, how can I compare the right side with the second statement you wrote? Is it legal? – McLovin Oct 27 '17 at 21:04
  • @Pilpel You assume-from the induction step-that $A>B$ . You want to show that $C>D$. But (by multiplying with $a+1$) you know that $A(a+1)=C$. Now if you only show that $B(a+1)>D$ you finish, hence the comparison. – MathematicianByMistake Oct 27 '17 at 21:33
1

multiplying the inequality $$(1+a)^k\geq 1+ka+\frac{k(k-1)}{2}a^2$$ by $1+a>0$ we get $$(1+a)^{k+1}\geq 1+a+ka(1+a)+\frac{k(k-1)(1+a)}{2}a^2$$ and it must be $$a+a(1+a)k-(k+1)a+\frac{k(k-1)(1+a)a^2}{2}-\frac{(k+1)ka^2}{2}>0$$ and this is true, since $$a^2k+\frac{ka^2}{2}\left(a(k-1)-2\right)>0$$

Dr. Sonnhard Graubner
  • 95,283