- #1
MathematicalPhysicist
Gold Member
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finding the sum of...
1*2+2*3+3*4+...+n(n+1)
i need to find to what it's equal?
i tried doing various ways:
2*(1+3)+4*(3+5)+6(5+7)+...
which means an even number times the sum of two odd numbers
so let's say:
a=2m
b=2m-1
a(b+3)+a(b+5)+a(b+7)...+n(n+1)
ab*n+a(3+5+7...+2n-1)*(n-1)=
a{[bn+(n-1)(3+5+7+...2n-1)]}
when 3+5+7+...(2n-1)=(3+s_n)n/2
when s_n=3+n(2n-1-3)=2n^2-4n+3
so (3+s_n)n/2=(2n^2-4n+6)n/2=n^3-2n^2+3n
here i pretty much given up, can someone help me.
1*2+2*3+3*4+...+n(n+1)
i need to find to what it's equal?
i tried doing various ways:
2*(1+3)+4*(3+5)+6(5+7)+...
which means an even number times the sum of two odd numbers
so let's say:
a=2m
b=2m-1
a(b+3)+a(b+5)+a(b+7)...+n(n+1)
ab*n+a(3+5+7...+2n-1)*(n-1)=
a{[bn+(n-1)(3+5+7+...2n-1)]}
when 3+5+7+...(2n-1)=(3+s_n)n/2
when s_n=3+n(2n-1-3)=2n^2-4n+3
so (3+s_n)n/2=(2n^2-4n+6)n/2=n^3-2n^2+3n
here i pretty much given up, can someone help me.