anonymous
  • anonymous
A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work. Sn: 12 + 42 + 72 + . . . + (3n - 2)2 = 𝑛(6𝑛2−3𝑛−1)2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
n(6n^2-3n-1)/2
anonymous
  • anonymous
anybody got a clue
anonymous
  • anonymous
@anonymous_user

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More answers

phi
  • phi
can you make this clearer (3n - 2)2 ?
anonymous
  • anonymous
n(6n^2-3n-1)/2
phi
  • phi
No, I mean the last term in the sum 12 + 42 + 72 + . . . + (3n - 2)2
anonymous
  • anonymous
its 1^2+4^2+7^2
anonymous
  • anonymous
im so lost
phi
  • phi
oh. it looks like twelve + forty-two+... and I was not making any progress (who would have guessed)
anonymous
  • anonymous
lol its my fualt i should have double checked thst sooo
phi
  • phi
Do you have to prove the result or derive it? proving it should be easier
anonymous
  • anonymous
no i just have to show thast the stamnets are true and show all my work proving them
phi
  • phi
are you studying proof by induction?
anonymous
  • anonymous
i think so
anonymous
  • anonymous
yes iam
phi
  • phi
the first step is show the identity is true for n=1 can you do that ?
anonymous
  • anonymous
no im jsut now learning this stug=ff lol sorry im a newbie
phi
  • phi
when n=1, what is "series" ?
phi
  • phi
what is the last term of the summation when n=1 ?
anonymous
  • anonymous
1
phi
  • phi
and the starting term is 1 in other words the entire summation is 1^2 (just 1 term) on the right-hand side we have the formula n(6n^2-3n-1)/2 what is that when n=1 ?
anonymous
  • anonymous
ummmmm
phi
  • phi
you sub in n=1 into the formula n(6n^2-3n-1)/2 what do you get ?
anonymous
  • anonymous
scratches head um n+1(6n^2-3n-1)/2
phi
  • phi
the idea is everywhere you see n in the formula, erase it, and write in 1 in its place then simplify (now that you have a number instead of a letter, you can do that) what do you get ?
anonymous
  • anonymous
1(61^2-31-1)/2
phi
  • phi
ok. and remember when you have for example 3n that means 3*n ( multiply signs are usually left out) so the formula is 1*(6*1^2 - 3*1 -1)/2 now simplify that
anonymous
  • anonymous
hold on im doing it now
anonymous
  • anonymous
the answer is 1 right
phi
  • phi
yes, it simplifies to (6-4)/2 = 2/2 = 1 that shows the formula works for n=1 because 1^2 = 1 (the answer we get from the formula)
phi
  • phi
now the hard part. we assume the formula works for numbers up to n (whatever n happens to be) and show it also works for n+1 This part is difficult because it looks like we need some messy algebra.
anonymous
  • anonymous
this is what i didi 1^2 evaluates to 1 Multiply 1 and 6 1 6*1^2 evaluates to 6 Multiply 1 and 3 1 3*1 evaluates to 3 6*1^2-3*1 evaluates to 3 6*1^2-3*1-1 evaluates to 2 Multiply 1 and 2 1 1*(6*1^2-3*1-1) evaluates to 2 1*(6*1^2-3*1-1)/2 evaluates to 1
anonymous
  • anonymous
uhhh math
anonymous
  • anonymous
so how dowe start this
phi
  • phi
yes, that is how you evaluate it.
anonymous
  • anonymous
sweet
phi
  • phi
all of that proves that the formula works for n=1 now assume the formula works for numbers up to n show it works for n+1 Induction Step: Assume the identity is true for n: \[ 1^2 + 4^2 + ... +(3n-1)^2 = \frac{n(6n^2-3n-1)}{2} \]
phi
  • phi
now add the next term to the summation that is add \( \left(3(n+1)-1 \right)^2 \) to both sides \[ 1^2 + 4^2 + ... +(3n-1)^2+\left(3(n+1)-1 \right)^2 = \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2\]
anonymous
  • anonymous
holy moly tat is hard
phi
  • phi
yes, it is very difficult.
anonymous
  • anonymous
so is there more too it plz say no lol
phi
  • phi
the formula for n+1 becomes \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \] we have to show that \[ \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2 \] can be written as \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \]
anonymous
  • anonymous
thnx

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