Let x + x 2 + x 3 + ⋯ + x n − 2 + x n − 1 + x n = x − 1 x ( x n − 1 ) . ( x = 1 )
What is 2 + 6 x + 1 2 x 2 + ⋯ + ( n 2 − 5 n + 6 ) x n − 4 + ( n 2 − 3 n + 2 ) x n − 3 + ( n 2 − n ) x n − 2 ?
A. ( x − 1 ) 4 ( n 2 − n ) x n + 2 − ( 2 n 2 − 2 ) x n + 1 + ( n 2 + n ) x n − 2
B. ( x − 1 ) 3 ( n 2 + n ) x n + 1 − ( 2 n 2 + 2 ) x n + ( n 2 − n ) x n − 1 + 2
C. ( x − 1 ) 3 ( n 2 − n ) x n + 1 − ( 2 n 2 − 2 ) x n + ( n 2 + n ) x n − 1 − 2
(Calculate and choose the answer carefully)
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Check my 'Geometric sequence and calculus (1)' :
1 + 2 x + 3 x 2 + . . . + ( n − 2 ) x n − 3 + ( n − 1 ) x n − 2 + n x n − 1 = ( x − 1 ) 2 ( x − 1 ) ( ( n + 1 ) x n − 1 ) − ( x n + 1 − x ) ( 1 )
= x 2 − 2 x + 1 n x n + 1 − n x n − x n + 1
Differentiate both sides w . r . t . x
2 + 6 x + 1 2 x 2 + ⋯ + ( n 2 − 5 n + 6 ) x n − 4 + ( n 2 − 3 n + 2 ) x n − 3 + ( n 2 − n ) x n − 2
= ( x − 1 ) 4 ( x − 1 ) 2 ( ( n 2 + n ) x n − n 2 x n − 1 − n x n − 1 ) − ( n x n + 1 − n x n − x n + 1 ) 2 ( x − 1 )
= ( x − 1 ) 3 ( n 2 − n ) x n + 1 − ( 2 n 2 − 2 ) x n + ( n 2 + n ) x n − 1 − 2
You can substitute some number into the equation to determine: Sub n = 3 , x = 4
LHS= ( 3 2 − 3 ( 3 ) + 2 ) 4 3 − 3 + ( 3 2 − 3 ) 4 3 − 2 = 2 + 6 ( 4 ) = 2 6
RHS= ( 4 − 1 ) 3 ( 3 2 − 3 ) 4 3 + 1 − ( 2 ( 3 2 ) − 2 ) 4 3 + ( 3 2 + 3 ) x 3 − 1 − 2 = 2 7 7 0 2 = 2 6