Question

the sum

Answer 1

lets say this question was asked before wolfram revolution....no calculator or wolframulator

Answer 2

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

Answer 3

sure...tag the math hater...that makes sense....

Answer 4

Is it 51???

Answer 5

yes its 51,thanks @sauravshakya this is nice work,without this is origininal without any computer...

Answer 6

Welcome

Answer 7

http://openstudy.com/users/jonask#/updates/508127d7e4b00774b241c4c3

Answer 8

can you see there formularsin math form or the funny form

Answer 9

i tried to re-write as\[\frac{ x^3 }{ x^3-(x-1)^3 }\]

Answer 10

\[\sum_{i=0}^{101}x^3=\frac{ 1 }{ 101^3 }(0+1^3+2^3+3^3+...+102^3)\]
\[\sum_{i=0}^{101}(x_i-1)^3=\frac{ -1 }{ 101^3 }(101^3+100^3+...+1^3)\]
together\[(\frac{ 1 }{ 101^3 }(1+102^3))\]

Answer 11

but is there a way to solve this approach

Answer 12

I dont think u can take the summation seperatly

Answer 13

so \[\sum_{}^{}\frac{ f(x) }{ g(x) }\neq \frac{ \sum_{}^{}f(x) }{ \sum_{}^{}g(x) }\]

Answer 14

YEP.

Answer 15

okay that sets it clear,fully solved problem thanks

Answer 16

makes sense

Answer 17

Welcome again.

Answer 18

\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }\neq \frac{ 2 }{ 5 }\]