Question

Question!

Answer 1

can i split \[\sum_{r=1}^{n} r^{5}\] to \[\sum_{r=1}^{n} r^{3} . \sum_{r=1}^{n} r^{2}\] ?

Answer 2

if not how should i evaluate \[\sum_{r=1}^{n} r^{5} \]

Answer 3

before answering that question, do you know what a sum notation represents ?

Answer 4

can you evaluate :

\[\large \sum \limits_{r=1}^5 r\]

?

Answer 5

yes

Answer 6

5

Answer 7

5r*

Answer 8

nope, \(r\) is just a dummy variable - its used only to index

Answer 9

is it 1+2+3+4+5 = 15

Answer 10

Yes !


\[\large \sum \limits_{r=1}^5 r\]

is same as

\[\large 1+2+3+4+5\]

Answer 11

\[\large \sum \limits_{r=1}^8 r^3\]

is same as

\[\large 1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3\]

Answer 12

i have the formula for r, r^2 and r^3

Answer 13

so you cannot split the sum as product of two other sums, but you can certainly split it as below :


\[\large \sum \limits_{r=1}^8 r^3 =\left(\sum \limits_{r=1}^8 r^3\right) + \left(\sum \limits_{r=1}^8 r^3\right) \]

Answer 14

Oh so are you trying to find a formula for fifth powers of first few numbers ?

Answer 15

so you cannot split the sum as product of two other sums, but you can certainly split it as below :


\[\huge \sum \limits_{r=1}^n r^5 = ?\]

Answer 16

hmm.. not necessarily the formula but how to evaluate it

Answer 17

is that what you're trying to find ?

Answer 18

yes

Answer 19

@ikram002p any idea how to find the fifth powers of first few numbers ?

Answer 20

wolfram gives this neat formula >.<
http://www.wolframalpha.com/input/?i=%5Csum+%5Climits_%7Br%3D1%7D%5En++r%5E5+

Answer 21

never derived it befor

Answer 22

show us how to work it now :)

Answer 23

like for 'r' it is \[\frac{n}{2}n(n+1)\]

Answer 24

erm.. but if i dont know the formula how to solve it ><

Answer 25

hahah nice im learning i never knew this stuff *0* sorry for butting in haha

Answer 26

Question is.. Conjecture a general formula for \[\frac{2}{n ^{3}}\sum_{r=1}^{n}(r ^{3}+3r ^{5})\], and prove it by mathematical induction

Answer 27

yes i know the proofs for below :

\[\large \begin{align} \\ \sum \limits_{r=1}^n r &= \dfrac{n(n+1)}{2} \\~\\
\sum \limits_{r=1}^n r^2 &= \dfrac{n(n+1)(2n+1)}{6} \\~\\
\sum \limits_{r=1}^n r^3 &= \left(\dfrac{n(n+1)}{2}\right)^2 \\~\\
\end{align}\]

Answer 28

never had to work beyond these haha!

Answer 29

wow !

Answer 30

yes. i was told to memorise up til r^3 too

Answer 31

i got\[\frac{2}{n ^{3}}[\sum_{r=1}^{n} r ^{3} +3 \sum_{r=1}^{n} r ^{5})]\]

Answer 32

ok where did everyone go.. =.=