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

- Mathsadness

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

- katieb

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- anonymous

sum as
\[6\sum_1^{30}k^2 +\sum_1^{30}1\]

- anonymous

second term is just 30

- anonymous

do you know the formula for \[\sum_{k=1}^n k^2\]?

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## More answers

- Mathsadness

no

- anonymous

its is \[\frac{n(n+1)(2n+1)}{6}\]

- anonymous

replace n by 30, compute and you will have the answer. i will write it if you like

- anonymous

\[6\times \frac{30(31)(71)}{6} +30\]
\[30(31)(71)+30\]

- anonymous

i get 66060

- Mathsadness

\[\sum_{k=1}^{n} k^3 \]
What is the formula for this??

- Mathsadness

...

- anonymous

that one is easy to remember.

- anonymous

do you know the one for
\[\sum_{k=1}^{n}k\]?

- anonymous

it is \[\frac{n(n+1)}{2}\]

- 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

so the square of Gauss' formula??

- anonymous

if that thing is called gauss's formula then yes

- Mathsadness

for the summation of n positive integers....

- anonymous

yes. here they are, more than you want to know
http://polysum.tripod.com/

- Mathsadness

wow!

- anonymous

oh yes, i see that it is called gauss' formula. i learn more here than in skule

- anonymous

if you like i can show you how to derive them, but you probably don't need to know it

- Mathsadness

how do you do that?

- Mathsadness

:P

- anonymous

ok we start with the first one. "gauss" and the method extends. you want
\[\sum_{k=1}^n k\]

- anonymous

write this
\[\sum_{k=1}^n k^2-(k-1)^2\]

- 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

so by observation we have \[\sum_{k=1}^n k^2-(k-1)^2=n^2\]

- anonymous

then do the algebra:
\[k^2-(k-1)^2 = k^2-(k^2-2k+1)=2k-1\]

- 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

now solve for
\[\sum_{k=1}^nk\] using algebra and you get \[\sum_{k=1}^n k = \frac{n(n+1)}{2}\]

- 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

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

cool!

- anonymous

problem 9 - 12 and and 20 - 23 work it out (not the algebra)

##### 1 Attachment

- anonymous

have fun

- Mathsadness

thx

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