Do you know how proof by induction works? I.e., the logic behind it and its process?
1. $5+10+15+...+5n=\frac{5n(n+1)}{2}$
2. $\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+...+\frac {4}{5^n}=1-\frac{1}{5^n}$
3. $2^n>n^2$, for $n \ge 5$, $n \varepsilon \mathbb{N}$
4. $5.6+5.6^2+2.6^3+....+5.6^n=6(6^n-1)$
i would appreciate any help so thanks in advance!
Hello, tinspire!
$\displaystyle 2.\;\;\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+ \text{ . . . }+\frac{4}{5^n}\:=\:1-\frac{1}{5^n}$
Verify $S_1\!:\;\;\dfrac{4}{5} \:=\:1- \dfrac{1}{5} \quad\Rightarrow\quad \dfrac{4}{5} \:=\:\dfrac{4}{5}$ . True!
Assume $S_k\!:\;\;\dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} + \text{ . . . }+\dfrac{4}{5^k} \;=\;1 - \dfrac{1}{5^k}$
Add $\dfrac{4}{5^{k+1}}$ to both sides:
$\qquad \dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} + \text{ . . . }+ \dfrac{4}{5^k} + \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{1}{5^k} + \dfrac{4}{5^{k+1}}\;\;[1]$
The right side is: $\:1 - \dfrac{1}{5^k} + \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{5}{5^{k+1}} + \dfrac{4}{5^{k+1}} \;=\; 1 - \dfrac{1}{5^{k+1}}$
Then [1] becomes: $\:\dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} +\text{ . . . }+ \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{1}{5^{k+1}}$
This is
The inductive proof is complete.
well let's look at the first one
let n=1
$P(1)=[5(1)=\dfrac {5(1)(1+1)}{2}]$
$5\overset{?}{=}5$
Yes. $5=5$ So $P(1)=True$
Now assume $P(n)=True$ and consider $P(n+1)$
$\displaystyle{\sum_{k=1}^{n+1}}5k=$
$5(n+1)+\displaystyle{\sum_{k=1}^{n}}5k=$
$5(n+1)+\dfrac {5n(n+1)}{2}=$
$\dfrac{2\cdot 5(n+1)+5n(n+1)}{2}=\dfrac{5(n+1)(n+2)}{2}=\dfrac{5 (n+1)((n+1)+1)}{2}$
and thus given the statement is true for $n$ it is true for $n+1$
You can work the 3 others. The key is to try and get the $P(n+1)$ expression to include $P(n)$ so you can substitute in the known simplification for $P(n)$.
3. 2^n > n^2, for n≥5, nεN
Base case:
n=5, 2^5=32 and 5^2= 25. Hence, lhs > rhs (Base case is true.)
Now, for any k > 5, assume that 2^k > k^2 is true.
Then, we shall prove that 2^(k+1) > (k+1)^2 is also true.
(But first, I am going to digress for a bit.)
2k^2 - (k+1)^2 = 2k^2 - k^2 - 2k -1 = k^2-2k - 1 = (k-1)^2 -2: since k > 5, it's clear that (k-1)^2 - 2 > 0.
My point? for any k > 5, 2k^2 - (k+1)^2 > 0, or better put, 2k^2 > (k+1)^2 starting from 5.
Then, 2^(k+1) = 2*2^k > 2* k^2 = 2k^2 > (k+1)^2 as needed.