Geometric sequence and calculus(2)

Calculus Level 3

Let x + x 2 + x 3 + + x n 2 + x n 1 + x n = x ( x n 1 ) x 1 x+x^2+x^3+\cdots+x^{n-2}+x^{n-1}+x^{n}=\dfrac{x(x^n-1)}{x-1} . ( x 1 x\neq1 )

What is 2 + 6 x + 12 x 2 + + 2+6x+12x^2+\cdots+ ( n 2 5 n + 6 ) x n 4 + ( n 2 3 n + 2 ) x n 3 + ( n 2 n ) x n 2 ? (n^2-5n+6)x^{n-4}+(n^2-3n+2)x^{n-3}+(n^2-n)x^{n-2} ?


A. ( n 2 n ) x n + 2 ( 2 n 2 2 ) x n + 1 + ( n 2 + n ) x n 2 ( x 1 ) 4 \frac{(n^2-n)x^{n+2}-(2n^2-2)x^{n+1}+(n^2+n)x^{n}-2}{(x-1)^4}

B. ( n 2 + n ) x n + 1 ( 2 n 2 + 2 ) x n + ( n 2 n ) x n 1 + 2 ( x 1 ) 3 \frac{(n^2+n)x^{n+1}-(2n^2+2)x^n+(n^2-n)x^{n-1}+2}{(x-1)^3}

C. ( n 2 n ) x n + 1 ( 2 n 2 2 ) x n + ( n 2 + n ) x n 1 2 ( x 1 ) 3 \frac{(n^2-n)x^{n+1}-(2n^2-2)x^n+(n^2+n)x^{n-1}-2}{(x-1)^3}

(Calculate and choose the answer carefully)

B C All of them are incorrect A

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1 solution

Tommy Li
May 29, 2016

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 1+2x+3x^2+...+(n-2)x^{n-3}+(n-1)x^{n-2}+nx^{n-1} = ( x 1 ) ( ( n + 1 ) x n 1 ) ( x n + 1 x ) ( 1 ) ( x 1 ) 2 \frac{(x-1)((n+1)x^n-1)-(x^{n+1} -x)(1)}{(x-1)^{2}}

= n x n + 1 n x n x n + 1 x 2 2 x + 1 = \frac{nx^{n+1}-nx^n-x^n+1}{x^2-2x+1}

Differentiate both sides w . r . t . x w.r.t. x

2 + 6 x + 12 x 2 + + ( n 2 5 n + 6 ) x n 4 + ( n 2 3 n + 2 ) x n 3 + ( n 2 n ) x n 2 2+6x+12x^2+\cdots+(n^2-5n+6)x^{n-4}+(n^2-3n+2)x^{n-3}+(n^2-n)x^{n-2}

= ( 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 ) 4 =\frac{(x-1)^2((n^2+n)x^n-n^2x^{n-1}-nx^{n-1})-(nx^{n+1}-nx^n-x^n+1)2(x-1)}{(x-1)^4}

= ( n 2 n ) x n + 1 ( 2 n 2 2 ) x n + ( n 2 + n ) x n 1 2 ( x 1 ) 3 =\frac{(n^2-n)x^{n+1}-(2n^2-2)x^n+(n^2+n)x^{n-1}-2}{(x-1)^3}


You can substitute some number into the equation to determine: Sub n = 3 , x = 4 n=3,x=4

LHS= ( 3 2 3 ( 3 ) + 2 ) 4 3 3 + ( 3 2 3 ) 4 3 2 (3^2-3(3)+2)4^{3-3}+(3^2-3)4^{3-2} = 2 + 6 ( 4 ) 2+6(4) = 26 26

RHS= ( 3 2 3 ) 4 3 + 1 ( 2 ( 3 2 ) 2 ) 4 3 + ( 3 2 + 3 ) x 3 1 2 ( 4 1 ) 3 = 702 27 = 26 \frac{(3^2-3)4^{3+1}-(2(3^2)-2)4^3+(3^2+3)x^{3-1}-2}{(4-1)^3} = \frac{702}{27} = 26

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