If #u_1,u_2,u_3,...# form a G.P with common ratio #k#, find the sum of #u_1u_2+u_2u_3+...+u_n u_(n+1)# in terms of #u_1# and #k#?

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Answer 1 · 2017-09-14T20:39:18

#u_1u_2+u_2u_3+...+u_n u_(n+1) = (ku_1^2(k^(2n)-1))/(k^2-1)#

Note that:

#(u_(k+1)u_(k+2))/(u_k u_(k+1)) = k^2#

So:

#u_1u_2+u_2u_3+...+u_n u_(n+1)#

is a geometric series with initial term:

#u_1u_2 = ku_1^2#

and common ratio #k^2#

The sum of a geometric series to #n# terms is given by the formula:

#sum_(k=1)^n ar^(n-1) = (a(r^n-1))/(r-1)#

where #a# is the initial term and #r# the common ratio.

So with initial term #a=ku_1^2# and common ratio #r=k^2#, we have:

#u_1u_2+u_2u_3+...+u_n u_(n+1) = (ku_1^2(k^(2n)-1))/(k^2-1)#