Mathematical Induction Problem

[Phyzwizz]

I was working on my homework and I did two of the mathematical induction problems before this one and they were super easy but I must be forgetting something because I just can't seem to solve this one.

2+7+12+17...+(5n-3) = (n/2)(5n-1)

So I know that step 1 is to prove that it works with 1, 2=(1/2)(5(1)-1) and yes 2=2
Then I know your supposed to "Assume that 2+7+12+17+...+(5n-3)=(n/2)(5n-1)".

This next part is where I get stuck...
I start off plugging in the (n+1) for n on the right side and then substituting (n/2)(5n-1) for 2+7+12+17 on the left side giving me... (n/2)(5n-1)+(5n-3)=((n+1)/2)(5(n+1)-1)
I tried multiplying everything out but that doesn't seem to get me anywhere.

Help would be greatly appreciated! Thanks.

 

[eumyang]

Compare to mine below, the part in bold

(n/2)(5n-1)+(5n-3)=((n+1)/2)(5(n+1)-1)

looks to be wrong.

The way I learned induction, I use another variable for part 2.

Assume true for n = k:
2 + 7 + 12 + 17 + ... + (5k-3) = \frac{k}{2}(5k-1)

Prove true for n = k + 1:
2 + 7 + 12 + 17 + ... + (5k-3) + (5k + 2)
= \frac{k}{2}(5k-1) + (5k + 2)

Add by finding the common denominator, and you should get to
\frac{(k + 1)}{2}(5k+4)

 

[Phyzwizz]

Okay thank you at least this points me back in the right direction. I was wondering how you got the (5k+2) on the right side.

And also I was wondering if the final part at the end of your work is the answer because it doesn't seem to end up being equal to the other side of the equation

 

[eumyang]

Okay thank you at least this points me back in the right direction. I was wondering how you got the (5k+2) on the right side.

It's actually the left side, simplified. Let me re-write it and maybe it will be more clear.

Prove true for n = k + 1:
\begin{aligned}<br /> 2 + 7 + 12 + 17 + ... + (5k-3) + (5(k+1) - 3) &amp;= \frac{(k + 1)}{2}(5(k+1) - 1) \\<br /> 2 + 7 + 12 + 17 + ... + (5k-3) + (5k + 2) &amp;= \frac{(k + 1)}{2}(5k + 4) \\<br /> \frac{k}{2}(5k-1) + (5k + 2) &amp;= \frac{(k + 1)}{2}(5k + 4)<br /> \end{aligned}

Now get the LHS equal to the RHS.

 

[HallsofIvy]

Okay thank you at least this points me back in the right direction. I was wondering how you got the (5k+2) on the right side.

The general term in the sequence is 5n- 3. Replacing n by n+ 1 gives 5(n+1)- 3= 5n+ 5- 3= 5n+ 2.

And also I was wondering if the final part at the end of your work is the answer because it doesn't seem to end up being equal to the other side of the equation

The final part is
\frac{k}{2}(5k- 1)+ 5k+ 2= \frac{k(5k- 1)+ 2(5k+ 2)}{2}
= \frac{5k^2- k+ 10k+ 4}{2}= \frac{5k^2+ 9k+ 4}{2}= \frac{(k+1)(5k+4)}{2}
Yes, they are the same.

 

[Ray Vickson]

I was working on my homework and I did two of the mathematical induction problems before this one and they were super easy but I must be forgetting something because I just can't seem to solve this one.

2+7+12+17...+(5n-3) = (n/2)(5n-1)

So I know that step 1 is to prove that it works with 1, 2=(1/2)(5(1)-1) and yes 2=2
Then I know your supposed to "Assume that 2+7+12+17+...+(5n-3)=(n/2)(5n-1)".

This next part is where I get stuck...
I start off plugging in the (n+1) for n on the right side and then substituting (n/2)(5n-1) for 2+7+12+17 on the left side giving me... (n/2)(5n-1)+(5n-3)=((n+1)/2)(5(n+1)-1)
I tried multiplying everything out but that doesn't seem to get me anywhere.

Help would be greatly appreciated! Thanks.

You are summing (5k - 3) over k from 1 to n, and sum(5k - 3) = 5*sum(k) - 3*sum(1).

RGV