Question

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

Answer 1

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

Answer 2

second term is just 30

Answer 3

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

Answer 4

no

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i get 66060

Answer 9

\[\sum_{k=1}^{n} k^3 \]

What is the formula for this??

Answer 10

...

Answer 11

that one is easy to remember.

Answer 12

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

Answer 13

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

Answer 14

the one for \[\sum_{k=1}^n k^3 \] is the square of that one. it is\[(\frac{n(n+1)}{2})^2\]

Answer 15

so the square of Gauss' formula??

Answer 16

if that thing is called gauss's formula then yes

Answer 17

for the summation of n positive integers....

Answer 18

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

Answer 19

wow!

Answer 20

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

Answer 21

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

Answer 22

how do you do that?

Answer 23

:P

Answer 24

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

Answer 25

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

Answer 26

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 \[n^2\]

Answer 27

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

Answer 28

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

Answer 29

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
\[2\sum_{k=1}^nk-\sum_{k=1}^n =n^2\]

Answer 30

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

Answer 31

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 \[\sum_{k=1}^n k^3-(k-1)^3=n^3\]

Answer 32

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

Answer 33

cool!

Answer 34

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

Answer 35

have fun

Answer 36

thx