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anonymous
 one year ago
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
anonymous
 one year ago
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

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phi
 one year ago
Best ResponseYou've already chosen the best response.1can you make this clearer (3n  2)2 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1No, I mean the last term in the sum 12 + 42 + 72 + . . . + (3n  2)2

phi
 one year ago
Best ResponseYou've already chosen the best response.1oh. it looks like twelve + fortytwo+... and I was not making any progress (who would have guessed)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol its my fualt i should have double checked thst sooo

phi
 one year ago
Best ResponseYou've already chosen the best response.1Do you have to prove the result or derive it? proving it should be easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i just have to show thast the stamnets are true and show all my work proving them

phi
 one year ago
Best ResponseYou've already chosen the best response.1are you studying proof by induction?

phi
 one year ago
Best ResponseYou've already chosen the best response.1the first step is show the identity is true for n=1 can you do that ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no im jsut now learning this stug=ff lol sorry im a newbie

phi
 one year ago
Best ResponseYou've already chosen the best response.1when n=1, what is "series" ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1what is the last term of the summation when n=1 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1and the starting term is 1 in other words the entire summation is 1^2 (just 1 term) on the righthand side we have the formula n(6n^23n1)/2 what is that when n=1 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1you sub in n=1 into the formula n(6n^23n1)/2 what do you get ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0scratches head um n+1(6n^23n1)/2

phi
 one year ago
Best ResponseYou've already chosen the best response.1the 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 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1ok. 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
 one year ago
Best ResponseYou've already chosen the best response.0hold on im doing it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer is 1 right

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, it simplifies to (64)/2 = 2/2 = 1 that shows the formula works for n=1 because 1^2 = 1 (the answer we get from the formula)

phi
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.0this 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*1^23*1 evaluates to 3 6*1^23*11 evaluates to 2 Multiply 1 and 2 1 1*(6*1^23*11) evaluates to 2 1*(6*1^23*11)/2 evaluates to 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how dowe start this

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, that is how you evaluate it.

phi
 one year ago
Best ResponseYou've already chosen the best response.1all 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: \[ 1^2 + 4^2 + ... +(3n1)^2 = \frac{n(6n^23n1)}{2} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now add the next term to the summation that is add \( \left(3(n+1)1 \right)^2 \) to both sides \[ 1^2 + 4^2 + ... +(3n1)^2+\left(3(n+1)1 \right)^2 = \frac{n(6n^23n1)}{2} +\left(3(n+1)1 \right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0holy moly tat is hard

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, it is very difficult.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is there more too it plz say no lol

phi
 one year ago
Best ResponseYou've already chosen the best response.1the formula for n+1 becomes \[ \frac{(n+1)}{2} \left(6(n+1)^2 3(n+1) 1\right) \] we have to show that \[ \frac{n(6n^23n1)}{2} +\left(3(n+1)1 \right)^2 \] can be written as \[ \frac{(n+1)}{2} \left(6(n+1)^2 3(n+1) 1\right) \]
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