Solved By Kyle Gooding
Expand the series on both sides of (1),(2) pg. 7-12b to verify these equalities. (1)
∑ j = 2 5 C j ∗ j ( j − 1 ) x j − 2 = ∑ j = 0 3 C j + 2 ∗ ( j + 2 ) ( j + 1 ) x j {\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) ∑ j = 1 5 C j ∗ j x j − 1 = ∑ j = 0 4 C j + 1 ∗ ( j + 1 ) x j {\displaystyle \sum _{j=1}^{5}C_{j}*jx^{j-1}=\sum _{j=0}^{4}C_{j+1}*(j+1)x^{j}}
Expanding both sides of (1) results in: ∑ 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=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}} ∑ 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 {\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: ∑ j = 2 5 C j ∗ j ( j − 1 ) x j − 2 = C 2 ( 2 ) + C 3 6 x + C 4 12 x 2 + C 5 20 x 3 {\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}} ∑ j = 0 3 C j ( j + 2 ) ( j + 1 ) x j = C 2 ( 2 ) + C 3 6 x + C 4 12 x 2 + C 5 20 x 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: ∑ j = 1 5 C j ∗ j x 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=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}} ∑ 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 {\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: ∑ j = 1 5 C j ∗ j x 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}} ∑ j = 1 5 C j ∗ j x 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}}
Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)
