Mathematical Induction's Please Help?

Question

Prove:
1+4+7+...+(3n-2)=1/2n(3n-1)

Assume for some K
1+4+7+...+(3k-2)=1/2k(3k-1)

Show true for K+1
???????????????

Answer 1

Start by proving that it is true for n = 1

f(n) = (1/2) * n * (3n - 1)
f(1) = (1/2) * 1 * (3(1) - 1)
f(1) = (1/2) * 1 * 2
f(1) = 1

So the sum of the first "1" numbers is 1, as you can see.

Now assume for some value k, you know that
f(k) = (1/2) * k * (3k - 1)

Prove it is true for f(k + 1):
f(k + 1) = f(k) + (3(k+1) - 2)
f(k + 1) = k(3k - 1)/2 + (3(k+1)- 2)
f(k + 1) = k(3k - 1)/2 + (3k + 3 - 2)
f(k + 1) = k(3k - 1)/2 + 2(3k + 1)/2
f(k + 1) = [k(3k - 1) + 2(3k + 1)] / 2
f(k + 1) = [3k^2 - k + 6k + 2] / 2
f(k + 1) = [3k^2 + 5k + 2] / 2
f(k + 1) = (k + 1)(3k + 2) / 2
f(k + 1) = (k + 1)(3(k + 1) - 1) / 2

Let n = k + 1
f(n) = n(3n - 1) /2

Since we've shown it true for n = 1. And assuming it is true for k, we have proven it true for k + 1. Therefore it is true for all n, by induction.

Answer 2

For this statement to be true for K+1, this would mean:
1+4+7+...+(3k - 2)+(3(k+1)-2) = 1/2*(k+1)(3(k+1)-1) = 1/2*(k+1)(3k+2)

Let's look at the right hand side:
[1+4+7+...+(3k - 2)]+(3(k+1)-2)
The terms in the brackets should look familiar: that's one side of your induction hypothesis! We've assumed that it equals 1/2k(3k-1)

Thus we have:
1+4+7+...+(3k - 2)+(3(k+1)-2)
= [1+4+7+...+(3k - 2)]+(3(k+1)-2)
=1/2*k(3k-1) + (3(k+1)-2)

Now it's just algebra. See if you can simplify this sum to show that it equals 1/2*(k+1)(3k+2)
Hint: multiply everything out, get a common denominator, then factor

One more thing: DON'T FORGET YOUR BASE STEP!
Remember to plug 1 in for n and make sure you get a true statement.
Good Luck!

Answer 3

Ah yes calculus...I'm studying that stuff for my midterm right now.

now say: (k+1) term = 3(k+1)-2
so that means that
S sub (k+1) is equal to 1/2k(3k-1)+2(k+1)-2 because that is the kth term plus the k+1 term.

Then simplify that and it should equal
1/2(k+1)(3(k+1)-1)
which proves the statement.

Answer 4

the left hand side is:
1+4+7+...+(3k-2) + 3(k+1)-2 = (1+4+7+...+(3k-2)) + 3(k+1)-2
= 1/2k(3k-1) + 3(k+1)-2 = 1/2 k (3k-1) + 3k+3-2 = 1/2 k (3k-1) + 3k+ 1

the right hand side is:
1/2(k+1)(3(k+1)-1) = 1/2 (k+1) ( 3k+3-1) = 1/2 (k+1) (3k+2)