Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n.
1. 2 + 4 + 6 + ... + 2n = n(n + 1)
4. 1 + 2 + 2^2 + ... + 2^(n-1) = 2^n - 1
2. 3 + 4 + 5 + ... + (n + 2) = 1/2n(n + 5)
5. 1 + 4 + 4^2 + ... + 4^(n-1) = 1/3(4^n - 1)
3. 11 + 23 + 35 + ... + (12n - 1) = n(6n + 5)
6. 1•2 + 2•3 + 3•4 + ... + n(n + 1) = 1/3n(n + 1)(n + 2)
Remember when n = 1 you are talking about the first term
1. 2 + 4 + 6 + ... + 2n = n(n + 1)
I. n=1 2(1) = 1(1 + 1)
2 = 2
II. Show: 2 + 4 + 6 + ... + 2k + 2(k + 1) = (k + 1)((k + 1) + 1) right side simplifies (k + 1)(k + 2)
Assume: 2 + 4 + 6 + ... + 2k = k(k + 1)
Now add the next term (by replacing k with (k + 1) in the preceding term) to the left side and use your algebra skills to show it matches the simplified form of the right side.
2 + 4 + 6 + ... + 2k = k(k + 1) Assumption
2 + 4 + 6 + ... + 2k + 2(k + 1)
k(k + 1) + 2(k + 1) Now you can either distribute, combine, factor. Or
(k + 1)(k + 2) recognize factor by grouping and factor out the (k + 1)
Notice this matches the simplified right side