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- unicwaan

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(n + 1)(8n + ) divided by 6

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- unicwaan

- schrodinger

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- campbell_st

there seems to be something missing from the sum (8n + ??)

- unicwaan

8n + 7 sorry

- campbell_st

ok.... so prove its true for n = 1
using the general term on the left you get 4 x 6 so the 1st term is divisible by 6
the sum of 1 term is( the right side) gives 4(1 + 1)(8x1 + 7) = 4 x 2 x 15
= 120
which seems to be a huge issue...1st term is 24 but the sum of 1 term is 120...
you might like to check my work....but it seems you already have a problem

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- Loser66

The problem is not clear!!

- campbell_st

and if you look at the sequence 5 x 7 = 35 not dividible by 6
so the sum of the 1st 2 terms won't be divible by 6

- Loser66

To me, I understand it as
\(4*6 +5*7 +6*8 +.......+4n(n+2)=\dfrac{4(n+1)(8n+7)}{6}\)
and prove it by induction

- unicwaan

yeah thats it

- campbell_st

makes a huge difference....
so test it with n = 1 what do you get on the left and right side...?

- Loser66

But to it, the basic step with n=1 is not true
since when n=1, the left hand side is \(4*1(4*1+2) = 4*6\) . the first element of the sequence.

- unicwaan

\[\frac{ 4(4n + 1)(8n + 7)}{ 6 }\] I forgot the other 4... im so sorry

- campbell_st

ok... so substitute n = 1 what do you get as the sum on the right side of the equation..?

- Loser66

while plug in n=1 to the right hand side, we get \(\dfrac{4(4*1+1)(8*1+7)}{6}= 50\)

- campbell_st

so it seems it fails the initial step of proof by induction

- unicwaan

the right side would equal 50... yeah what @Loser66 just said. and I dont understand why n equals 1

- campbell_st

I'm sure in the question you are given an initial condition that n = 1, 2, 3, ...
but if it's not stated the normal process is to start with n = 1 and prove the left hand side is equal to the right had side
so n = 1 is really saying find the 1st term.... and show what the sum of 1 term is...

- campbell_st

some induction questions use sigma notation

- unicwaan

So if the sum of the left side (when plugging in 1) doesn't equal the first term of the right side, then the statement is false?

- campbell_st

the left side is the sum of the terms....with the last term being the general case
the right side is the general case for the sum of the terms....
so the initial task is to prove the 1st term is equal to the sum of 1 term...
it might seem trivial... but it is an essential part of the process
in your question things failed the initial test so there is no need to go further.

- unicwaan

Okay, that makes, thank you so much!!

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