Preform a mathematical induction proof

Question

1•12+2•13+3•14+...+n(n+11)=n(n+1)(n+17) / 3

Answer 1

In order to do an induction proof, there are always two parts, and it helps to think of it like a row of dominoes. First you knock over the first domino, which is the base case. Then you show that ANY domino falling knocks over the next one, which is the inductive step. Together those two facts mean every domino falls.

First let us be clear about the general term. Looking at the pattern 1 times 12, then 2 times 13, then 3 times 14, the second factor is always 11 more than the first, so the nth term is n(n + 11). We are proving:

1(12) + 2(13) + ... + n(n + 11) = n(n + 1)(n + 17)/3

BASE CASE, n = 1. The left side is just the first term, 1 times 12 = 12. The right side is:

1(2)(18)/3 = 36/3 = 12

Both sides give 12, so the statement holds for n = 1.

INDUCTIVE STEP. Assume the formula works for some n = k. That assumption is the inductive hypothesis:

1(12) + 2(13) + ... + k(k + 11) = k(k + 1)(k + 17)/3

Now we have to show it forces the formula to work for k + 1. The sum up to k + 1 is just the sum up to k plus one more term, and that next term is (k + 1)(k + 12), since we plug k + 1 into n(n + 11):

sum to (k+1) = k(k + 1)(k + 17)/3 + (k + 1)(k + 12)

Here is where the inductive hypothesis earned its keep. We replaced the whole long sum with the closed formula, and now it is just algebra.

Notice that (k + 1) is a common factor in both pieces, so we factor it out. Putting everything over 3:

= (k + 1)[k(k + 17) + 3(k + 12)]/3

Expanding inside the brackets:

= (k + 1)[k^2 + 17k + 3k + 36]/3

= (k + 1)(k^2 + 20k + 36)/3

Remember what we are aiming for. Replacing n with k + 1 in the original formula should give (k + 1)(k + 2)(k + 18)/3, so we want that quadratic to factor into (k + 2)(k + 18). Checking, 2 times 18 = 36 and 2 + 18 = 20, so it does:

= (k + 1)(k + 2)(k + 18)/3

That is exactly the formula with k + 1 in place of n.

So, our final answer is that the statement holds for n = 1, and whenever it holds for k it also holds for k + 1, so by mathematical induction it is true for all positive integers n.

Answer 2

Base step: n = 1

1(1+11) = 1(1+1)(1+17)/3

12 = 12

This proves the base step.

Induction step: Assume the assertion true for n and check validity for n+1.

1.12 + ... + n(n+11) + (n+1)(n+12) = (using induction hypothesis)

n(n+1)(n+17)/3 + (n+1)(n+12) =

(n+1)(n(n+17)/3 + n+12) =

(n+1)(n² + 17n + 3n + 36)/3 =

(n+1)(n+2)(n+18)/3

The last line is the right hand side of the given assertion after replacing

n with n+1.

That proves the assertion.