For any positive integer n, prove that (1+1/n)^n < (1+1/n+1)^n+1
Answers
Step-by-step explanation:
Show that for n ≥ 1, we have
(a) Xn
k=1
k =
n(n + 1)
2
(b) Xn
k=1
(2k − 1) = n
2
.
Solution. 1
(a): We first check the base case, n = 1. The sum is 1 and the formula evaluates to
1 as well, so we are good.
Induction hypothesis: Assume that for some n ≥ 1 we have
Xn
k=1
k =
n(n + 1)
2
.
We add (n + 1) on both sides of this relation and get
nX
+1
k=1
k = (n + 1) + n(n + 1)
2
.
The right hand side can now be rewritten as
nX
+1
k=1
k =
[n + 1]([n + 1] + 1)
2
.
Thus, we have proved that if the formula holds for n, it holds for n + 1. By the
principle of mathematical induction, the identity is true for all integers n ≥ 1.
(b): We first check the base case, n = 1. Both sides evaluates to 1, so we are ok.
Induction hypothesis: Assume that for some n ≥ 1 we have
Xn
k=1
(2k − 1) = n
2
.
We add 2(n + 1) − 1 on both sides of this relation and get
nX
+1
k=1
(2k − 1) = 2(n + 1) − 1 + n
2
.
The right hand side can now be rewritten as
nX
+1
k=1
(2k − 1) = [n + 1]2
.
Thus, we have proved that if the formula holds for n, it holds for n + 1. By the
principle of mathematical induction, the identity is true for all integers n ≥ 1.