Question

compute the sum

Answer 1

\[\sum_{i=0}^{101}\frac{ x _{i}^3 }{ 1-3x _{i} +3x _{i}^2}\]
\[x _{i}=1/101\]

Answer 2

\[x _{1}=1/101\]

Answer 3

I think you're missing some rule for xi in terms of i

Answer 4

imeant x1 not xi

Answer 5

x1=1/101

Answer 6

the solution is quite easy

Answer 7

is your answer supposed to be in terms of xi?
because you haven't included a rule for x in terms of i

Answer 8

oh i see its\[x _{i}=\frac{ i }{ 101 }\]

Answer 9

does that help

Answer 10

\[\frac{ -x^3 }{ (x-1)^3-x^3 }\]

Answer 11

yeah

Answer 12

from here i thinthe denominator is easy to work with cos two consecutiver cubes cancel

Answer 13

or we can work out
\[\frac{ 1 }{ 1-(\frac{ x-1 }{ x })^3 }\]

Answer 14

\[\frac{ 1 }{ 1-(1-\frac{ 1 }{ x })^3 }\]

Answer 15

\[\sum_{i=0}^{101}=(1-1/x _{i})=?\]

Answer 16

error

Answer 17

\[\sum_{i=0}^{101}x _{i}^{3}=\frac{ 1 }{ 101^3 }(0+1^3+2^3+3^3+4^3+5^3+...+102^3)\]
\[\sum_{i=0}^{101}(x_{i}-1)^3= \frac{ 1 }{ 101^3 }((-1)^3+0+1^3+2^2+... +101^3)\]
difference

Answer 18

\[=\frac{ 1 }{ 101^3 }(1+102^3)\]

Answer 19

\[\frac{ (0+1^3+2^3+3^3+4^3+...+102^3) }{ 1+102^3 }\]

Answer 20

wolfram=51
i have no idea

Answer 21

@hartnn can you pls help

Answer 22

i made mistake with \[(x-1)^3\]
sum

Answer 23

\[\sum_{i=0}^{101}(x_i-1)^3=\frac{ -1 }{ 101^3 }(101^3+100^3+99^3+...0^3+1^3)\]

Answer 24

sum\[(\frac{ 2 }{ 101^3})(0+1+3^3+...+101^3)+\frac{ 102^3 }{ 101^3 }=\frac{ 2(0+1+3^3+...+101^3)+102^3 }{ 101^3 }\]

Answer 25

total
\[\frac{ 0+1+3^3+...+102^3 }{2(0+1+1+3^3+... 101^3 )+102^3}\]

Answer 26

unsloved

Answer 27

solved