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# Show that ∑(n+ r-1)(n+r) = (1/3n)(7n^2 – 1)?

Show that ∑(n+ r-1)(n+r) = 1/3n(7n^2 – 1)

Find ∑1/(n+ r-1)(n+r) in terms of n

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- lou hLv 71 decade agoBest Answer
∑ _{r=1, n} 1= n

∑ _{r=1, n} r= (n+1)·n/2

∑ _{r=1, n} r²= n·(n+1)·(2n+1)/6

==> ∑ _{r=1, n} (n+ r-1)(n+r) = ∑ _{r=1, n} (n²+2nr +r² -n -r) =

(n²-n)·∑ _{r=1, n} 1 + (2n-1)· ∑ _{r=1, n} r + ∑ _{r=1, n} r² =

(n²-n)·n + (2n-1)·(n+1)·n/2 + n·(n+1)·(2n+1)/6 =

n/6 ·[ 6·(n²-n)+3·(2n-1)·(n+1) +(n+1)·(2n+1)] =

n/6· [6n²-6n+6n²+3n-3+2n²+3n+1] =

n/6· [14n²-2] = 1/3· n·(7n²-1)

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1/[(n+r-1)·(n+r)]= 1/(n+r-1) - 1/(n+r) ==>

∑ _{r=1, n} 1/[(n+r-1)·(n+r)]= ∑ _{r=1, n} [1/(n+r-1) - 1/(n+r) ] =

[ 1/n - 1/(n+1) ] + [ 1/(n+1) - 1/(n+2)] +....+ [1/(2n-1)- 1/(2n)] =

1/n - 1/(2n) = 1/(2n)

Saludos

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