Mathsadness
  • Mathsadness
sum{k=1 to 30} (6k^2+1) How would you find the value of this sum?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
sum as \[6\sum_1^{30}k^2 +\sum_1^{30}1\]
anonymous
  • anonymous
second term is just 30
anonymous
  • anonymous
do you know the formula for \[\sum_{k=1}^n k^2\]?

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Mathsadness
  • Mathsadness
no
anonymous
  • anonymous
its is \[\frac{n(n+1)(2n+1)}{6}\]
anonymous
  • anonymous
replace n by 30, compute and you will have the answer. i will write it if you like
anonymous
  • anonymous
\[6\times \frac{30(31)(71)}{6} +30\] \[30(31)(71)+30\]
anonymous
  • anonymous
i get 66060
Mathsadness
  • Mathsadness
\[\sum_{k=1}^{n} k^3 \] What is the formula for this??
Mathsadness
  • Mathsadness
...
anonymous
  • anonymous
that one is easy to remember.
anonymous
  • anonymous
do you know the one for \[\sum_{k=1}^{n}k\]?
anonymous
  • anonymous
it is \[\frac{n(n+1)}{2}\]
anonymous
  • anonymous
the one for \[\sum_{k=1}^n k^3 \] is the square of that one. it is\[(\frac{n(n+1)}{2})^2\]
Mathsadness
  • Mathsadness
so the square of Gauss' formula??
anonymous
  • anonymous
if that thing is called gauss's formula then yes
Mathsadness
  • Mathsadness
for the summation of n positive integers....
anonymous
  • anonymous
yes. here they are, more than you want to know http://polysum.tripod.com/
Mathsadness
  • Mathsadness
wow!
anonymous
  • anonymous
oh yes, i see that it is called gauss' formula. i learn more here than in skule
anonymous
  • anonymous
if you like i can show you how to derive them, but you probably don't need to know it
Mathsadness
  • Mathsadness
how do you do that?
Mathsadness
  • Mathsadness
:P
anonymous
  • anonymous
ok we start with the first one. "gauss" and the method extends. you want \[\sum_{k=1}^n k\]
anonymous
  • anonymous
write this \[\sum_{k=1}^n k^2-(k-1)^2\]
anonymous
  • anonymous
it 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
  • anonymous
so by observation we have \[\sum_{k=1}^n k^2-(k-1)^2=n^2\]
anonymous
  • anonymous
then do the algebra: \[k^2-(k-1)^2 = k^2-(k^2-2k+1)=2k-1\]
anonymous
  • anonymous
in other words there was no square in it. now we know that \[\sum_{k=1}^n 2k-1=n^2\] 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
  • anonymous
now solve for \[\sum_{k=1}^nk\] using algebra and you get \[\sum_{k=1}^n k = \frac{n(n+1)}{2}\]
anonymous
  • anonymous
f 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 \[\sum_{k=1}^n k^3-(k-1)^3=n^3\]
anonymous
  • anonymous
do 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
Mathsadness
  • Mathsadness
cool!
anonymous
  • anonymous
problem 9 - 12 and and 20 - 23 work it out (not the algebra)
anonymous
  • anonymous
have fun
Mathsadness
  • Mathsadness
thx

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