University of Florida/Egm4313/s12.team11.gooding/R3

Given

${\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=\sum _{j=0}^{3}C_{j+2}*(j+2)(j+1)x^{j}}$

(2)
${\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=\sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}}$

Solutions

Expanding both sides of (1) results in:
${\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)(2-1)x^{0}+C_{3}(3)(3-1)x^{1}+C_{4}(4)(4-1)x^{2}+C_{5}(5)(5-1)x^{3}}$
${\displaystyle \sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{0+2}(0+2)(0+1)x^{0}+C_{1+2}(1+2)(1+1)x^{1}+C_{2+2}(2+2)(2+1)x^{2}+C_{3+2}(3+2)(3+1)x^{3}}$

Simplifying:
${\displaystyle \sum _{j=2}^{5}C_{j}*j(j-1)x^{j-2}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}}$
${\displaystyle \sum _{j=0}^{3}C_{j}(j+2)(j+1)x^{j}=C_{2}(2)+C_{3}6x+C_{4}12x^{2}+C_{5}20x^{3}}$

 The two sums are equal.

Expanding both sides of (2) results in:
${\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)x^{0}+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}$
${\displaystyle \sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}=C_{0+1}(0+1)x^{0}+C_{1+1}(1+1)x^{1}+C_{1+2}(2+1)x^{2}+C_{2+2}(3+1)x^{3}+C_{3+2}(4+1)x^{4}}$

Simplifying:
${\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}$
${\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=C_{1}(1)+C_{2}(2)x^{1}+C_{3}(3)x^{2}+C_{4}(4)x^{3}+C_{5}(5)x^{4}}$

 The two sums are equal.

Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)