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[slapstick4235]

Can someone explain how this:
E_n=E_n-1+1/3(n-1)³+n²-1/3n³
becomes this?
En=E_n-1+n-1/3

 

[stapel]

Can someone explain how this:
E_n=E_n-1+1/3(n-1)³+n²-1/3n³
becomes this?
En=E_n-1+n-1/3

What is the definition of \(\displaystyle E_n\)? Also, are your equations as follows?

. . . . .\(\displaystyle E_n\, =\, E_{n-1}\, +\, \dfrac{1}{3(n\, -\, 1)^3}\, +\, n^2\, -\, \dfrac{1}{3n^3}\)

. . . . .\(\displaystyle E_n\, =\, E_{n-1}\, +\, n\, -\, \dfrac{1}{3}\)

Thank you!

 

[lookagain]

Can someone explain how this:
E_n=E_n-1+1/3(n-1)³+n²-1/3n³
becomes this?
En=E_n-1+n-1/3

slapstick4235, you should put grouping symbols around the fractional coefficients:

E_n = E_n-1 + (1/3)(n - 1)³ + n² - (1/3)n³

E_n = E_n-1 + n - 1/3

- - - - - - -- - - -- -- - - -- - - - -- - - -

Expand \(\displaystyle \ (n - 1)^3.\)


\(\displaystyle (n - 1)^3 \ = \)

\(\displaystyle (n - 1)(n - 1)(n - 1) \ =\)

\(\displaystyle (n - 1)(n^2 - 2n + 1) \ = \)

\(\displaystyle n(n^2 - 2n + 1) \ - \ 1(n^2 - 2n + 1) \ =\)

\(\displaystyle n^3 - 2n^2 + n - n^2 + 2n - 1 \ = \)

\(\displaystyle n^3 - 3n^2 + 3n - 1\)


Substitute this back in, multiply through by 1/3, and combine like terms.


E_n = E_n-1 + (1/3)(n^3 - 3n^2 + 3n - 1) + n² - (1/3)n³


Finish it from here.