Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

Answer 1

how much have you progressed ?

Answer 2

check it for 1 , and then for k , and then for k+1

Answer 3

You essentially have two steps to complete:
1. Show that it is true for n=1
2. Show that if it is true for k, then it is true for k+1
These combine to form the inductive proof!

Answer 4

I checked for 1 for [n(6n^2 - 3n - 1)] / 2 and got 16... Does this mean that the statement is false?

Answer 5

its coming 1 only i think u have done the calculation wrong it is true for 1

Answer 6

(3n-2)^2 is true for 1

Answer 7

OH

Answer 8

For 1 you would get 1^2 = 1 on the LHS
and 1*(6*1-3*1-1)/2 = 2/2 = 1 on the RHS
Since the LHS = RHS, it is true for n=1.

Answer 9

How do I get 6k^3 + 9k^2 -7k +13 to equal to [k+1(6(k+1)^2 -3(k+1)-1]/2 ? In other words, LHS = RHS ?

Answer 10

Sorry for the late response..
Right for the second step what you want to do is say "suppose it is true for some integer k"
That would mean that \[\sum_{m=1}^k{(3m-2)^2}=\frac{k(6k^2-3k-1)}{2}\]Remember that this is only a supposition.
Now you say what happens if you add the (k+1)th term to each side..
\[\sum_{m=1}^k{(3m-2)^2}+(3(k+1)-2)^2=\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2\]You can take the term on the LHS into the sum and tidying up the RHS gets you:
\[\sum_{m=1}^{k+1}{(3m-2)^2}=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]This matches the identity for n=k+1 right?
This means that if the identity is true for k, then it is true for k+1.
Since it is true for n=1, using what we just worked out, it is then true for 2.
Since it is true for n=2, it is then true for 3.
We could continue this forever, so it is true for all positive integers.

The hardest part is tidying up the RHS.