For n = 1; 2; 3;... , let:
L(n) =INT(between 1&0) of (x^(n+1))/(2 - x)dx.
By writing x^n = x^(n+1)(2 - (2 - x)), show that this sequence of numbers satises the recurrence
relation: L(n+1) = 2L(n)-1/n. For n = 1; 2; 3;... , let:
L(n) =INT(between 1&0) of (x^(n+1))/(2 - x)dx.
By writing x^n = x^(n+1)(2 - (2 - x)), show that this sequence of numbers satises the recurrence
relation: L(n+1) = 2L(n)-1/n. @Mathematics

Question

For n = 1; 2; 3;... , let:
L(n) =INT(between 1&0) of (x^(n+1))/(2 - x)dx.
By writing x^n = x^(n+1)(2 - (2 - x)), show that this sequence of numbers satises the recurrence
relation: L(n+1) = 2L(n)-1/n. For n = 1; 2; 3;... , let:
L(n) =INT(between 1&0) of (x^(n+1))/(2 - x)dx.
By writing x^n = x^(n+1)(2 - (2 - x)), show that this sequence of numbers satises the recurrence
relation: L(n+1) = 2L(n)-1/n. @Mathematics