Use mathematical induction to prove that 5+9+13+17+...+(4n+1)=n(2n+3). Help Please!

Math problem help?? Not sure how to go about this one.

Well, when n = 1, the left hand side is 4(1) + 1 = 5 and the right hand side is (1)(2(1) + 3) = 5. So your equation is true when n = 1.

Suppose that the statement is true when n = k for some integer k ≥ 1. That is, assume that

5 + 9 + .. + (4k+1) = k(2k + 3).

Now, consider the case n = k + 1. Using the above induction hypothesis we get

5 + 9 + ... + (4k + 1) + (4(k+1)+1) = k(2k + 3) + 4k + 5 = 2k² + 7k + 5 =

= (k + 1)(2k + 5) = (k + 1)(2(k + 1) + 3).

This is the required result for the case n = k + 1. We can conclude that the statement is true for all integers n ≥ 1.

First prove it true for n = 1. Done.

Now, show if that if is true for case n, then it must be true for case n+1.

So n(2n+3) = sigma (4n+1) starting at one.

we add the n+1 term 4(n+1) +1 = 4n+5

n(2n+3) =2n^2 +3n

If we add the next term, it becomes 2n^2 +7n +5, (adding the two terms together)

(n+1)(2(n+1) +3) = (n+1)(2n+2 +3) = (n+1)(2n+5) = 2n^2 +2n +5n +5 = 2n^2 +7n +5. I took the expression and evaluated it for n+1, and showed that it was the same as adding the (n+1)th term to the expression. QED.

STEP 1, prove for any value of n:

n=1: lhs = 5 i.e. 1-term, rhs = n(2n+3) = 1(5) = 5 so proven for n=1.

STEP 2, assume works for a given value k =n:

lhs= 5 + 9 + 13 +.... (4k + 1)
n=k
∑ (4n+1) = k(2k+3)
n=0

STEP 3, now try n = k+1 and see if still holds:

lhs = 5 + 9 + 13 + .... (4k+4+1) = {5 + 9 + 13 + ....+(4k+1) + 4(k+1)+1}
= k(2k+3) + 4k+5 by terms from step 2,
= 2k^2 +3k + 4k +5 = 2k^2 + 7k + 5 = (k +1)(2k+5)
= (k+1)[2(k+1) + 3]
= rhs for n =k+1 .... 3STEPS <=> QED.