The partial sum is sn = (3n)/(n+1).
With guess and check, I found that the answer is an = 3/((n)(n+1))
It looks like you have to some reverse work with telescoping series, but I don't know how to find the answer with just the partial sum provided.
With guess and check, I found that the answer is an = 3/((n)(n+1))
It looks like you have to some reverse work with telescoping series, but I don't know how to find the answer with just the partial sum provided.
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You can use the formula S(n+1)-S(n)=a(n+1)
So (3(n+1)/((n+1)+1)) - (3n)/(n+1) = a(n+1)
simplified (3n+3)/(n+2) - (3n)/(n+1) = a(n+1)
plug in (n-1) now for n, so that the left side will equal a(n)
(3(n-1)+3)/((n-1)+2) - (3(n-1))/((n-1)+1) = a(n)
simplify: (3n)/(n+1) - (3n-3)/n = a(n)
cross-multiply and multiply factors: 3n^2-(3n^2+3n-3n-3)/(n(n+1)) = a(n)
simplify: 3/(n(n+1)) = a(n)
Science.
So (3(n+1)/((n+1)+1)) - (3n)/(n+1) = a(n+1)
simplified (3n+3)/(n+2) - (3n)/(n+1) = a(n+1)
plug in (n-1) now for n, so that the left side will equal a(n)
(3(n-1)+3)/((n-1)+2) - (3(n-1))/((n-1)+1) = a(n)
simplify: (3n)/(n+1) - (3n-3)/n = a(n)
cross-multiply and multiply factors: 3n^2-(3n^2+3n-3n-3)/(n(n+1)) = a(n)
simplify: 3/(n(n+1)) = a(n)
Science.
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