Question

Use mathematical induction to prove the following proposition:
P(n) : 3 + 5 + 7 + ….+2n+ 1 = n(2 + n) where n = 1, 2, 3, ….

Answer 1

Ok, long time... But I think I got it right:
We have the following statement
\[
P(n): 3 + 5 + 7 + ... + (2n + 1) = n(2 + n) \\
n \ge 1
\]
Now, we define P(k) this way. It is indeed the same.
\[
P(k): 3 + 5 + 7 + ... + (2k + 1) = k(2 + k) \\
\]
Now we claim this to be true:
\[
P(k) = P(k-1) + (2 \cdot k + 1)
\]
So like, that the sum of all the numbers to (k) is sum of P(k-1) + item_of_k.
Lets prove it:
\[
k(2 + k) = (k-1) \cdot \bigg[\;2 + (k-1) \; \bigg] + (2k+1) \\
2k + k^2 = 2(k-1) + (k-1)(k-1) + 2k+1 \\
2k + k^2 = 2k-2 + k^2 - 2k + 1 + 2k+1 \\
2k + k^2 = 2k + k^2 \\
0 = 0

\]
Since we got an equality we can say that our claim was true.
Problem is that our claim is always based on sum of previous items...
For the first item (n = 1) we can't do this because there are no previous items.
So we just have to check if it works manually..
\[
\text{if n = 1:} \\
P(1) = 2 \cdot 1 + 1 \\
1 \cdot (2 + 1) = (2 \cdot 1 + 1) \\
3 = 3
\]
Which is also true.
And finally since it's all right, we can say:
\[
n(2 + n) = (n-1) \cdot \bigg[\;2 + (n-1) \; \bigg] + (2n+1) \\
\]

Answer 2

Thank you