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# Evaluate ##sum (2n+1)/2^n## from n=0 to infinity ?

6

The given expression can be split in to two series ##sum_0^oo (2n)/2^n + sum_0^oo 1/2^n ##. The second series is a with first term as 1 and common ratio ##1/2## and as such its sum would be ##1/(1-1/2)## =2

For the first series consider function f(x)= ##1/(1-x)## which is represented by the geometric series ##1+x+x^2+x^3+... ## , where x is bounded as -1 < x <1.

Now ##f' (x)= 1/ (1-x)^2 = 1+ 2x +3x^2 +....##.

If ##x= 1/2## it would become ##1/(1-1/2)^2= 1+ 2/2 +3/2^2+...n/2^(n-1)+... ## This means 4= ##sum_0^oo n/2^(n-1) = sum_0^oo (2n)/2^n ##

##sum_0^oo (2n)/2^n + sum_0^oo 1/2^n ##= 4+2 =6