Question

PLEASE HELP WILL GIVE MEDAL
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6
prove that n=1

Answer 1

what are the dots ?

Answer 2

all you really have to do is plug 1 into all of the n's .

Answer 3

4 time 6 , mulitply

Answer 4

well start by testing for n = 1


and show the 1st term is equal to the sum of 1 terms....

i.e. the left hand side equals the right hand side

Answer 5

okay

Answer 6

4(4(1)+1)(8(1)+7)/6

Answer 7

@campbell_st can walk you through this i am sure
i just want to ask if you understand how to do a proof by induction, i.e. what the steps are
after understanding what you need to do, there is only a bit of algebra to do it

Answer 8

well I think the problem fails for n = 1

so it can't be proven

Answer 9

not quite

Answer 10

how would the problem be false

Answer 11

did you get the sum of 1 term, n = 1, as 50..?

Answer 12

i have 4x1(4x1+2)=4x(4+2)=4x6

Answer 13

=24

Answer 14

and the 1st term is 4.6 which is 24..

so the sum of 1 term doesn't equal the 1st term...

Answer 15

so all you can do is show that the sum of the terms is false as it fails at n = 1

Answer 16

@melchar do you understand the how to do induction...?

Answer 17

not really

Answer 18

ok... its a 3 step process

step 1: prove the formula is true for the smallest value, which is usually n = 1

so is the 1st term equal to the sum of 1 terms

which means you substitute n = 1 into the right hand side of the equation

in you question you found the sum of 1 term was 50 while the 1st term was

24. So it couldn't be proven

Step 2: Assume that the formula is true for sum integer, n = k

so in your question you would have said the sum of k terms is

\[s_{k} =\frac {4(4k +1)(8k + 7)}{6}\]

Step 3: Prove that the formula is true for the term n = k+1

Which means you need to show the sum to k plus the (k +1) term
is equal to the sum of k + 1 terms.

\[S_{k +1} = S_{k} + a_{k + 1}\]

so you need to use the assumed sum, from step 2 and add the next term

This is the algebra part

\[\frac{4(4(k + 1) + 1)(8(k + 1)+7)}{6} = \frac{4(4k + 1)(8k + 7)}{6} + 4(k + 1)(4(k + 1 + 2)\]

hope that makes some sense

Answer 19

here are some notes from Sydney University on doing mathematical induction

http://sydney.edu.au/stuserv/documents/maths_learning_centre/induction.pdf

Answer 20

okay thank you very much !!! really helped