Question

when using induction for example if i had to prove say 2 +5 + ... + 3n -1 = n/3 ( n+2).. if it were true would i add n +1 to both sides or n+1 to every n i see?

Answer 1

n+1 to every n you see

Answer 2

The general approach is:

1. Show its true for some base case (in this case you would show its true for n=1)
2. Assume its true for n=k
3. Then prove it must also be true for n=k+1

Answer 3

ok thank you i have a similar problem just wanted to know how to start it after i prove P1 is true.... just so i understand 3n -1 + n+1 = ((n+1)/3)((n+1)+2)

Answer 4

not quite - let me show you step-by-step...

Answer 5

you are told to prove that:\[2 +5 + ... + (3n -1) = n( n+2)/3\]
1, Show this is true for n=1
For n=1 we know the last term in the sum is (3*1-1)=(3-1)=2
So the left-hand-side of your identity becomes 2
The right-hand-side for n=1 becomes 1(1+2)/3=3/3=1

In this case the left-hand-side and right-hand-sides do not match therefore this identity is false.

Answer 6

I think your identity should be:\[2 +5 + ... + (3n -1) = n( 3n+1)/2\]

Answer 7

then, for n=1, the right-hand-side evaluates to 1(3*1+1)/2=4/2=2 which matches the left-hand-side.

we can now move onto to step 2...

Answer 8

2. Assume its true for n=k

This gives us:\[2+5+...+(3k-1)=k(3k+1)/2\]

Now we move to step 3...

Answer 9

yeah i understand how to prove whether it's true or not...but how do start to prove Pn+1...... so 2 + 5+ ... + (3n-1) +n+1 = n(3(n+1)+1)/2 is this what i need to show when proving it on the right side

Answer 10

3. Then prove it must also be true for n=k+1

This gives us (on the left-hand-side):\[2+5+...+(3(k+1)-1)\]

Answer 11

we now need to manipulate this expression so that we end up with something that looks like our step 2...

Answer 12

couldn''t you just right it 2+5+...(3k-1) +(k+1)

Answer 13

\[2+5+...+(3(k+1)-1)=2+5+...+(3k-1)+(3(k+1)-1)\]what I did here was include the term before the last one

Answer 14

now you should spot that all terms up to \((3k-1)\) match our identity in step 2 (which we assumed was true)

Answer 15

therefore:\[2+5+...+(3(k+1)-1)=2+5+...+(3k-1)+(3(k+1)-1)\]\[=k(3k+1)/2+(3(k+1)-1)=k(3k+1)/2+(3k+3-1)\]\[=k(3k+1)/2+3k+2\]

Answer 16

why induction for something so simple.. but okay

Answer 17

your question about "couldn''t you just right it 2+5+...(3k-1) +(k+1)" is not correct.

the identity states:\[2+5+...+(3n-1)=n(3n+1)/2\]this hold for "n" terms.
so, if we have "k" terms, we get:\[2+5+...+(3k-1)=n(3k+1)/2\]and if we have "k+1" terms we get:\[2+5+...+(3(k+1)-1)=(k+1)(3(k+1)+1)/2\]

Answer 18

@dan815 - I believe the asker wants to learn about induction

Answer 19

sorry I forgot to replace one "n" with "k" above ^^^

Answer 20

@fiberdust - does this make sense now?

Answer 21

basically you have to replace EVERY "n" with "k+1" to get the correct identity for the k+1'th case

Answer 22

yeah so add k+1 where u see an n every time?

Answer 23

not "add" but replace every "n" with "k+1"

Answer 24

ok thank you all for your help and thoughts for understanding the process better with this type of problem

Answer 25

To take a very simple example, lets say you wanted to add all integers up to n, you would have:\[1+2+...+n\]so the sum of all integers up to k would be:\[1+2+...+k\]and the sum of all integers up to k+1 would be:\[1+2+...+(k+1)=1+2+...+k+(k+1)\]

Answer 26

yw

Answer 27

is there something wrong,. with the starting equation

Answer 28

yes - the identity was not stated correctly

Answer 29

oh okay wow ya

Answer 30

wasttinng my time!!