Just checking my answers.

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May 12th 2008, 07:55 PM
Tally

Just checking my answers.

a1 = 3, ak+1 = ak − 2

What is the value of the 4th term? (Select one.)

-9
-5
-3
1
43

I selected -3 for this. I am pretty sure about this.

F(n) = 2 + 4 + 6 +...+ 2n

Which expression below represents the given function? (Select one.)

F(n) = n(n + 1)

F(n) = (n − 1)(n + 1)

F(n) = n(n − 1)

F(n) = 2n(2n + 1)

F(n) = n2 + 2n

I guessed at answer 4, so could do with some help if I got it wrong.
May 12th 2008, 08:16 PM
Aryth

Could you explain the function:

F(n) = n2 + 2n

Is it 2n + 2n Or:

$F(n) = n^2 + 2n$

Or something else?
May 12th 2008, 10:42 PM
Moo

Hello,

Quote:

Originally Posted by Tally

a1 = 3, ak+1 = ak − 2

What is the value of the 4th term? (Select one.)

-9
-5
-3
1
43

I selected -3 for this. I am pretty sure about this.

Ok for this.

Quote:

F(n) = 2 + 4 + 6 +...+ 2n

Which expression below represents the given function? (Select one.)

F(n) = n(n + 1)

F(n) = (n − 1)(n + 1)

F(n) = n(n − 1)

F(n) = 2n(2n + 1)

F(n) = n2 + 2n

I guessed at answer 4, so could do with some help if I got it wrong.

No it's not.

Note that your sequence is $2(1+2+\dots +n)$

What's in brackets is equal to $\frac{n(n+1)}{2}$ as sum of the first n consecutive numbers.

Hence $F(n)=2 \cdot \frac{n(n+1)}{2}=n(n+1)$ (Wink)
May 13th 2008, 03:46 AM
Tally

Thank you for your reply. I am still having trouble with Induction. For some reason this is a long old road.(Doh)

Could you explain why where the /2 came from as I don't see it in the function.
May 13th 2008, 07:33 AM
Aryth

$\sum_{i=1}^n 2i = 2\sum_{i=1}^n i$

Where:

$\sum_{i=1}^n i = \frac{n(n + 1)}{2}$

So:

$2\frac{n(n + 1)}{2} = n(n + 1)$

We have to verify it for the first case:

$P(1) = 1(1 + 1) = 2$

Now we make our assumption:

$P(k) = 2 + 4 + 6 + ... + 2k = k(k + 1)$

Now we have to show that:

$P(k + 1) = 2 + 4 + 6 + ... + 2k + 2(k + 1) = (k + 1)((k + 1) + 1)$

$= P(k) + 2(k + 1)$

$= k(k + 1) + 2(k + 1)$

$= k^2 + k + 2k + 2$

$= k^2 + 3k + 2$

$= (k + 1)(k + 2)$

$= (k + 1)((k + 1) + 1)$

Therefore the form $n(n + 1)$ is valid for all $n \in \mathbb{N}$