mathematical induction?

Question

k i have P_(n)=> Summation sign with n above it and i=1 underneath it with (4i-1) = 2n^2+n (n>or equal to 1)
i have proved it for n=1, assumed its also true for n=k but having trouble proving it for P_(k+1)...........
thanks if you can help

Answer 1

These are fun. But hard to explain. Especially on the internet where it can be hard to use the symbols. But ill try and give you the gist.

So you can assume P(n). If you rewrite your sum but this time you need to go from 1 to n+1. But now, I want you to write out the first few terms of the sum, and then do the normal ... bit inbetween and finish off with the nth and n+1th term.

On the other side of the equality you need to rewrite your formula for the summation replacing n by n+1

Now look on the LHS. Notice anything? You have a long sum from 1 to n, but you assume that this equals your handy nth term formula. So now you have on your LHS, a nice formula with the nth term, plus the extra n+1 term. On the RHS you have your formula but with n+1 rather than n. Now you need to rearrange to get them the same.

TIP: Sometimes this rearranging can be a real cruncher. Sometimes its a good idea to pick n as some random number [I like to use Pi] and work out both sides numerically with a calculator. If they are the same you have most likely got it right. Then you know that the manipulation you are doing is actually going to get you there. You can also work on both sides at the same time and meet in the middle, although if you do this it is more attractive to get back to the original form so that your LHS is the same as what your RHS has always been.

This is a bit wordy, but hopefully you're issue was with the method not the manipulation so I have helped a little.

Answer 2

Prove by induction.

n
Σ(4i - 1) = 2n² + n
i=1

If we plug in n = 1

4*1 - 1 = 2*1² + 1
4 - 1 = 2 + 1
3 = 3
So it is true for n = 1.

Now assume it is true for n = k and prove it is also true
for n = k + 1.

(2k² + k) + [4(k + 1) - 1] = 2k² + k + 4k + 4 - 1

= (2k² + 4k + 2) + (k + 1) = 2(k² + 2k + 1) + (k + 1)

= 2(k + 1)² + (k + 1)

So it is true for k + 1 also.

Therefore it is proved by induction.

Answer 3

summation from i=1 to n :(4i -1)
= 4 summation from i=1 to n (i) -summation from i=1 to n (1)
= 4 *n/2 *(n+1) - n
= 2n^2 + 2n -n
= 2n^2 +n
Using mathematical induction :
At n=1
L.H.S = 3
R.H.S = 2*1+1 = 3
So the the statment is true at n = 1
Assuming it is true at n =k
So : R.H.S = 2k^2 + k
At n = k+1
L.H.S =2k^2+k+(4(k+1 -1)=2k^2+5k+3 =(2k^2 + 4k + 2) +k+1
= 2(k^2+2k+1)+k+1
= 2(k+1)^2 + (k+1) = R.H.S
So the statment is true for all values

Answer 4

sum[k+1] = sum[k] + (4(k+1)-1)
= 2k^2+k + 4k+3
= 2(k+1)^2 +(k+1)
so if true for k, also true for k+1

Answer 5

k
∑ (4i-1) = 2k²+k


k+1
∑ (4i-1)
=
k
∑ (4i-1) + [4(k+1)-1]
=
{2k²+k} + {4k+3}
=
2k²+5k+3

But also:

k+1
∑ (4i-1) = 2(k+1)²+(k+1)

= 2(k²+2k+1) + (k+1)
= 2k²+5k+3


(Both the same)