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Powers of Natural Numbers Series[3]

Powers of natural numbers series can also be found algebraically by making use of the power of a binomial.

First n Natural Numbers Series[1]

For the first n natural number series, 1+2+3+...+n, a binomial to the power of 2 can be used. By making use of the identity  (n+1)2=n2+2n+1⇒(n+1)2-n2=2n+1. Let n=n-1 and substitute into the identity, then (n)2-(n-1)2=2(n-1)+1. By repeating the n=n-1 substitution, and get (n-1)2-(n-2)2=2(n-2)+1. By repeating the n=n-1 substitution again, and get (n-2)2-(n-3)2=2(n-32=2(n-3)+1. By continuing the n=n-1 substitution process until the original variable n is reduced to 1, that is ..., ((n-(n-2))+1)2-(n-(n-2))2=2(n-(n-2))+1, (1+1)2-(1)2=2(1)+1,  a totally n sets of identities are obtained. By adding up all the identities, the left hand side is reduced to the first and last term, (n+1)2-12, while the right hand side remains the summation of all terms. After rearranging the similar terms on the right hand, the identity becomes (n+1)2-12=2(n+(n-1)+(n-2)+(n-3)+...+2+1)+n(1). Imply (n+(n-1)+(n-2)+(n-3)+...+2+1)=(((n+1)2-1)-n)/2=(n2+n)/2=n(n+1)/2. Imply

First n Natural Number Squares Series[3]

For the first n natural number squares series, 12+22+32+...+n2, a binomial to the power of 3 can be used. By making use of the identity  (n+1)3=n 3+ 3n2+3n+1⇒(n+1)3-n3=3n2+3n+1. Let n=n-1 and substitute into the identity, then (n)3-(n-1)3=3(n-1)2+3(n-1)+1. By repeating the n=n-1 substitution, and get (n-1)3-(n-2)3=3(n-2)2+3(n-2)+1. By repeating the n=n-1 substitution again, and get (n-2)3-(n-3)3=3(n-3)2+3(n-3)+1. By continuing the n=n-1 substitution process until the original variable n is reduced to 1, that is ..., (n-(n-2)+1)3-(n-(n-2))3=3(n-(n-2))2+3(n-(n-2))+1, (1+1)3-(1)3=3(1)2+3(1)+1,  a totally n sets of identities are obtained. By adding up all the identities, the left hand side is reduced to the first and last term, (n+1)3-13, while the right hand side remains the summation of all terms. After rearranging the similar terms on the right hand, the identity becomes (n+1)3-13=3(n2+(n-1)2+(n-2)2+(n-3)2+...+22+12)+3(n+(n-1)+(n-2)+(n-3)+...+2+1)+n(1). Imply (n2+(n-1)2+(n-2)2+(n-3)2+...+22+12) =((n+1)3-1-3(n+(n-1)+(n-2)+(n-3)+...+2+1)-n)/3=(n3+3n2+3n+1-1-n)/3-(n+(n-1)+(n-2)+(n-3)+...+2+1) =(n3+3n2+2n)/3-n(n+1)/2 =(2(n3+3n2+2n)-3(n2+n))/6 =(2n3+3n2+n)/6=(n(2n2+3n+1))/6=(n(n+1)(2n+1))/6. Imply

First n Natural Number Cubes Series[3]

For the first n natural number cubes series, 13+23+33+...+n3, a binomial to the power of 4 can be used. By making use of the identity  (n+1)4=n 4+4n3+6n2+4n+1⇒(n+1)4-n4= 4n3+6n2+4n+1. Let n=n-1 and substitute into the identity, then (n)4-(n-1)4= 4( n-1) 3+6 ( n-1) 2+4 (n-1)+1. By repeating the n=n-1 substitution, and get (n-1)4-(n-2)4= 4 (n-2) 3+6 (n-2) 2+4 ( n-2)+1. By repeating the n=n-1 substitution again, and get (n-2)4-(n-3)4= 4 ( n-3) 3+6 (n-3) 2+4 ( n-3)+1. By continuing the n=n-1 substitution process until the original variable n is reduced to 1, that is ..., (n-(n-2)+1)4-(n-(n-2))4= 4(n-(n-2))3+6(n-(n-2))2+4(n-(n-2))+1, (1+1)4-(1)4=4 (1)3+6(1)2+4(1)+1,  a totally n sets of identities are obtained. By adding up all the identities, the left hand side is reduced to the first and last term, (n+1)4-14, while the right hand side remains the summation of all terms. After rearranging the similar terms on the right hand, the identity becomes (n+1)4-14= 4(n3+(n-1)3+(n-2)3+(n-3)3+...+23+13)+6(n2+(n-1)2+(n-2)2+(n-3)2+...+22+12)+4(n+(n-1)+(n-2)+(n-3)+...+2+1)+n(1). Imply (n3+(n-1)3+(n-2)3+(n-3)3+...+23+13) =((n+1)4-1-6(n2+(n-1)2+(n-2)2+(n-3)2+...+22+12)-4(n+(n-1)+(n-2)+(n-3)+...+2+1)-n)/4=(n 4+4n3+6n2+4n+1-1-n)/4-6(n2+(n-1)2+(n-2)2+(n-3)2+...+22+12)/4-(n+(n-1)+(n-2)+(n-3)+...+2+1) =(n 4+4n3+6n2+3n)/4-n(n+1)(2n+1)/4-n(n+1)/2 =((n4+4n3+6n2+3n)- (2n3+2n 2+ n2+n) -2(n2+n))/4 =( n4+2n3+n2)/4=(n 2(n2+2n+1))/4=(n 2(n+1) 2)/4=((n(n+1))/2) 2. Imply