Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

Answer 1

@satellite73

Answer 2

replace 4n by n, that should be a bit simpler.

Answer 3

can u show me? @experimentX

Answer 4

type that Question here.

Answer 5

it is a raft of algebra

Answer 6

you get to assume
\[4\times 6+5\times 7+..+4k(4k+2)=\frac{4(4k+2)(8k+7)}{6}\]

Answer 7

then replace \(k\) by \(k+1\) and see you can get the left hand side to equal to the right hand side

Answer 8

the tricky part is writing down the right hand side
the left hand side is
\[4\times 6+5\times 7+...+4k(4k+2)+4(k+1)(4(k+1)+2)\] and notice that is it
\[\overbrace{4\times 6+5\times 7+...+4k(4k+2)}+4(k+1)(4(k+1)+2)\]

Answer 9

the part under the braces we get to replace by expression we had on the right

Answer 10

so it becomes
\[\frac{4(4k+2)(8k+7)}{6}+4(k+1)(4(k+1)+2)\] on the left
of course you need algebra to clean this up a great deal

Answer 11

hmm ... didn't you miss something. between 4k and 8k+2 and (4(k+1))(8(k+2) + 2)

Answer 12

on the right hand side if you replace \(k\) by \(k+1\) you get
\[\frac{4(4(k+1)+2)(8(k+1)+7)}{6}\]

Answer 13

and you job it to show those two things are equal
@experimentX maybe, what did i miss?

Answer 14

unlike n and n+1 there are few numbers between 4n and 4n + 4
4n(4n+2), (4n + 2)(4n+4), (4n + 4)(4n + 6)
------------ this might be missing
just a guess. I usually make misatake

Answer 15

no i don't think i am missing anything. the next term after the kth term is the k + 1 st term

Answer 16

the number pattern seems like
(3+n)(n+5) though the ending is given as of the form 4n(4n+2) ... i was just guessing there might be some numbers between 4n(4n+2) and 4(n+1)(4(n+1) + 2)

Answer 17

this is tough to prove for all numbers because it fails the base case of n = 1

Answer 18

\[\large 4[1](4[n]+2)=\frac{4(4[1]+1)(8[1]+7)}{6}\]
\[\large 4(6)=\frac{4(5)(15)}{6}\]
\[24=50\]