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# Some more proof by induction help plzz! watch

1. 1). Prove by induction, for all positive integers n, that:

(1 x 5) + (2 x 6) + (3 x 7) + ..... + n(n + 4) = 1/6 n(n+1)(2n+13)

2). A sequence U1, U2, U3.....Un is defined by Un+1 = (3Un - 1)/4

The first term (U1) = 2

Prove by induction that Un = 4 [(3/4)^n] - 1

Help?!?
2. what have you done so far?
3. for question 2,

ive proved it for n = 1,

and now I have to prove it for n + 1

so ive just written it out:

4 [(3/4)^n+1] - 1

But what do I do now?
4. anyone?
5. (Original post by DeanK2)
what have you done so far?
any help ?
6. You need to be careful in setting out proofs by induction. So it's true for n = 1, and hence for some n = k, .

So , and substitute in place of U_k.
7. which one do you need help with?
8. 4 [(3/4)^n+1] - 1 = Un+1

so now separte Un+1 into Un + (n+1) ie the nth term + the n+first term

so
Un+1 = Un + (n+1)
= 4 [(3/4)^n] - 1 + 4 [(3/4)] - 1

BUT!!!!!!!! check the n+1st term... tis been so long since i've done induction that could well be wrong!!!
from then on its clear sailing, the trick is jsut to separate Un+1 into Un + (n+1)
9. 1). Prove by induction, for all positive integers n, that:

(1 x 5) + (2 x 6) + (3 x 7) + ..... + n(n + 4) = 1/6 n(n+1)(2n+13)

With induction you assume P(n), and prove p(1) p(n+1)

You prove that this:

1/6 n(n+1)(2n+13) + (n+1)((n+1) + 4)

is equal to this:

1/6 (n+1)((n+1)+1)(2(n+1)+13)

Explanation:

Your given a sequence and told the nth term is "n(n + 4)", and your told the sum of the first n terms is "1/6 n(n+1)(2n+13)". You need to prove that the sum of the first n terms + the (n+1)th term is equal to the sum of the first (n+1) termss.

Hope that helps

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