evaluate \\sum_{k=1}^infinity (k/2^k) evaluate \\sum_{k=1}^infinity (k^2/2^k) hi

Question

evaluate sum_{k=1}^infinity (k/2^k)
evaluate sum_{k=1}^infinity (k^2/2^k)


hint: to do this, you should take the square of the geometric series sum_{k=0}^infinity (1/2^k). The square of the series sums to 4 (since the series itself sums to 2), but if you evaluate the product series, you get sum_{k=0}^infinity (k+1)/2^k. Thus sum_{k=0}^infinity converges to 4; from this you can work out what sum_{k=1}^infinity converges to and (multiplying again by sum_{k=0}^infinity) what sum_{k=1}^infinity k^2/2^k converges to.

Explanation / Answer

S =k/2^k
S = 1/2 + 2/4 + 3/8 + 4/16+.................+infinite terms............(1)
multipying both sides by 1/2
S/2 = 1/4 + 2/8 + 3/16 + 4/32+.........(2)
solving (1) - (2)
S/2 = 1/2 + 1/4 + 1/8 + 1/16+...........
this forms a GP whose first term,a = 1/2 &common ratio, r = 1/2
S/2 = a/(1-r)
S/2 = 1/2/[1-1/2]
S/2 = 1
S = 2
i hope u can do the 2nd part now...applying same method

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