#u_1,u_2,u_3#,... are in Geometric progression(GP).The common ratio of the terms in the series is K.Now determine the sum of the series #u_1u_2+u_2u_3+u_3u_4+...+u_n u_(n+1)# in the form of K and #u_1#?

1 Answer
George C.
Jun 26, 2017

#sum_(k=1)^n u_k u_(k+1) = (u_1^2K(1-K^(2n)))/(1-K^2)#

Explanation:

The general term of a geometric progression can be written:

#a_k = a r^(k-1)#

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

The sum to #n# terms is given by the formula:

#s_n = (a(1-r^n))/(1-r)#

#color(white)()#
With the information given in the question, the general formula for #u_k# can be written:

#u_k = u_1 K^(k-1)#

Note that:

#u_k u_(k+1) = u_1 K^(k-1) * u_1 K^k = u_1^2 K^(2k-1)#

So:

#sum_(k=1)^n u_k u_(k+1) = sum_(k=1)^n u_1^2 K^(2k-1)#

#color(white)(sum_(k=1)^n u_k u_(k+1)) = sum_(k=1)^n (u_1^2 K)*(K^2)^(k-1)#

#color(white)(sum_(k=1)^n u_k u_(k+1)) = sum_(k=1)^n a r^(k-1)" "# where #a=u_1^2K# and #r = K^2#

#color(white)(sum_(k=1)^n u_k u_(k+1)) = (a(1-r^n))/(1-r)#

#color(white)(sum_(k=1)^n u_k u_(k+1)) = (u_1^2K(1-K^(2n)))/(1-K^2)#

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