unicwaan
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
campbell_st
  • campbell_st
there seems to be something missing from the sum (8n + ??)
unicwaan
  • unicwaan
8n + 7 sorry
campbell_st
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Loser66
  • Loser66
The problem is not clear!!
campbell_st
  • 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
  • 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
  • unicwaan
yeah thats it
campbell_st
  • campbell_st
makes a huge difference.... so test it with n = 1 what do you get on the left and right side...?
Loser66
  • 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
  • unicwaan
\[\frac{ 4(4n + 1)(8n + 7)}{ 6 }\] I forgot the other 4... im so sorry
campbell_st
  • campbell_st
ok... so substitute n = 1 what do you get as the sum on the right side of the equation..?
Loser66
  • 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
  • campbell_st
so it seems it fails the initial step of proof by induction
unicwaan
  • unicwaan
the right side would equal 50... yeah what @Loser66 just said. and I dont understand why n equals 1
campbell_st
  • 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
  • campbell_st
some induction questions use sigma notation
unicwaan
  • 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
  • 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
  • unicwaan
Okay, that makes, thank you so much!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.