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Question

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experimentx · Answer

put x=1 on f(x+1) = f(x)+3(x+1)+1
find f(2), f(3), ... up to few terms using inductive method.

experimentx · Answer

you get the succeeding value from preceding value

anonymous · Answer

@experimentX: Are you alluding Interpolation?

experimentx · Answer

until we find pattern @FoolForMath know a lot better when i comes to sensing sequence.

experimentx · Answer

i don't know the exact term ...

experimentx · Answer

what going to happen if f'(x+1) = f'(x) + 3 ??

experimentx · Answer

slope increasing at a constant rate.

anonymous · Answer

f(x) = 7

anonymous · Answer

This is a nice problem guys.

experimentx · Answer

never done this kinda problem before.

anonymous · Answer

1, 8, 18, 31,47,...

anonymous · Answer

I got a solution by using the recurrence:
$$f(x+1)=f(x)+3(x+1)+1$$ You can change this to:
$$f(x)=f((x-1)+1)=f(x-1)+3x+1$$.  But for f(x-1) we get:$$f(x-1)=f((x-2)+1)=f(x-2)+3(x-1)+1$$So for f(x) we obtain:$$f(x)=f(x-2)+3(x-1)+1+3x+1$$

anonymous · Answer

Those terms are consistent to:$$ \frac 32 x^2 + \frac 52 x − 3 $$

anonymous · Answer

If you continue this until you reach 1, you get:$$f(x)=f(1)+\sum_{i=2}^{x}(3i+1)$$

anonymous · Answer

Seems like I did it!! Yay me :D

Btw I took some help from here.

anonymous · Answer

http://answers.yahoo.com/question/index?qid=20110830164955AAEBzG2

anonymous · Answer

since f(1)=1, and:$$\sum_{i=2}^{x}(3i+1)=\frac{3n}{2}(n+1)+n-4$$, you'll obtain what fool for math posted.

experimentx · Answer

whoa ... what a question.

anonymous · Answer

Really good one!

anonymous · Answer

oops.  instead of n, i should have put x's.  Its weird to think of x as an integer.