Question

help with summation problem

Answer 1

\[\sum_{k=1}^{6}(k-k^2)\]

Answer 2

That's not too long; you can simply expand the summation and do it manually in this case...

Answer 3

the given problem can be solved as:
\[\sum_{k=1}^{6}k - \sum_{k=1}^{6}k ^{2}\]
the summation can be broken down as
\[((k)(k+1)/2) - ((k)(k+1)(k+2)/6)\] where k=6
this implies the summation solves to 49

Answer 4

i just used the property of additon of natural numbers and addition of their squares

Answer 5

the sum is -70

Answer 6

sorry thats not "49" thats "-49"

Answer 7

\[\sum_{k=1}^{6}(k-k^2)=\sum_{k=1}^{6}k(1-k)\]

\[=1(0)+2(-1)+3(-2)+4(-3)+5(-4)+6(-5)=-70\]

Answer 8

where did i do the mistake @zarkon please figure it out

Answer 9

\[\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 10

ohh i got my mistake......... thanks sir