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anonymous
 2 years 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.
Sn: 12 + 42 + 72 + . . . + (3n  2)2 = (n(〖6n〗^23n1))/2
anonymous
 2 years 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. Sn: 12 + 42 + 72 + . . . + (3n  2)2 = (n(〖6n〗^23n1))/2

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0just @sloths this is my question do u see this

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0gimme a second i'll look

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0this question doesn't make any sense. not sure if its the wording or copy paste wasn't good for it

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0its 1^2 and 4^2 not 12 or 42

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0u want me to send u a snapshot or attach the actual file

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0either would be great

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0k ill attach the file

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yep, which number is it?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i'm sorry my computer keeps crashing

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh thx no problem, u can always emailo me at msm6205@gmail.com if u cant help me here

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0were u able to look at the problem

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@heroromen11 hey u can help

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0plus i really don't understand this.. i'm not in pre calc yet

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh thx for trying though

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0can u help me with any other problem in the packet
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