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Mathsadness
 5 years ago
sum{k=1 to 30} (6k^2+1) How would you find the value of this sum?
Mathsadness
 5 years ago
sum{k=1 to 30} (6k^2+1) How would you find the value of this sum?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sum as \[6\sum_1^{30}k^2 +\sum_1^{30}1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0second term is just 30

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the formula for \[\sum_{k=1}^n k^2\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its is \[\frac{n(n+1)(2n+1)}{6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0replace n by 30, compute and you will have the answer. i will write it if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[6\times \frac{30(31)(71)}{6} +30\] \[30(31)(71)+30\]

Mathsadness
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{n} k^3 \] What is the formula for this??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that one is easy to remember.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the one for \[\sum_{k=1}^{n}k\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is \[\frac{n(n+1)}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the one for \[\sum_{k=1}^n k^3 \] is the square of that one. it is\[(\frac{n(n+1)}{2})^2\]

Mathsadness
 5 years ago
Best ResponseYou've already chosen the best response.0so the square of Gauss' formula??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if that thing is called gauss's formula then yes

Mathsadness
 5 years ago
Best ResponseYou've already chosen the best response.0for the summation of n positive integers....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes. here they are, more than you want to know http://polysum.tripod.com/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yes, i see that it is called gauss' formula. i learn more here than in skule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you like i can show you how to derive them, but you probably don't need to know it

Mathsadness
 5 years ago
Best ResponseYou've already chosen the best response.0how do you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok we start with the first one. "gauss" and the method extends. you want \[\sum_{k=1}^n k\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0write this \[\sum_{k=1}^n k^2(k1)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it might not be obvious, but the part after the sigma is one term minus the previous one. so this is a telescoping sum, and when you are done (it will be obvious if you write out some terms) everything will be killed off except the \[n^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so by observation we have \[\sum_{k=1}^n k^2(k1)^2=n^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then do the algebra: \[k^2(k1)^2 = k^2(k^22k+1)=2k1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in other words there was no square in it. now we know that \[\sum_{k=1}^n 2k1=n^2\] because the part after the sum was the same. this is equal to \[2\sum_{k=1}^nk\sum_{k=1}^n =n^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now solve for \[\sum_{k=1}^nk\] using algebra and you get \[\sum_{k=1}^n k = \frac{n(n+1)}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f course you have probably seen this proved a different way. but now the fun begins. if you want the formal for \[\sum_{k=1}^n k^2 \] you write \[\sum_{k=1}^n k^3(k1)^3=n^3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do the algebra and the \[k^3\] will drop ouit (just like the k^2 did before) and you will get a summation only involving \[k^2\] and k and 1. you already have the one for k, so solve for \[\sum_{k=1}^nk^2\] and you will get your answer. this trick works all the way up, but the algebra gets really really annoying so best just to have a list

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0problem 9  12 and and 20  23 work it out (not the algebra)
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