How do you use sigma notation to write the sum for #1/4+3/8+7/16+15/32+31/64#?

2 Answers
Feb 19, 2017

Answer:

# 1/4+3/8+7/16+15/32+31/64 = sum_(r=1)^5 (2^n-1)/2^(n+1) #

Explanation:

If we look at the sequence for the terms in denominator we have:

# {4,8,16,32,64}#

It should be obvious that these numbers are successive powers of #2#, with the first term #2# missing so the general term of the above sequence would be:

# u_n=2^(n+1) # where #n in {1,2,3,4,5}#

And now we look at the sequence for the terms in the numerator

# {1,3,7,15,31} #

and note that the terms are one less than successive powers of #2#, so the general term of this sequence be:

# u_n=2^n-1 # where #n in {1,2,3,4,5}#

Hence a general term of the series would be:

# u_n=(2^n-1)/2^(n+1)# where #n in {1,2,3,4,5}#

And hence we can write the finite series using sigma notation as

# 1/4+3/8+7/16+15/32+31/64 = sum_(r=1)^5 (2^n-1)/2^(n+1) #

NB: The sum evaluates to #129/64#

Feb 19, 2017

Answer:

#sum_(n=1)^5 (2^n-1)/(2^(n+1))#

Explanation:

Note that:

#1/4 = (2^1-1)/2^2#

#3/8 = (2^2-1)/2^3#

etc...

so the sum is:

#sum_(n=1)^5 (2^n-1)/(2^(n+1))#

We can also write it as:

#S = sum_(n=1)^5 (2^n-1)/(2^(n+1)) = sum_(n=1)^5 (1/2-1/(2^(n+1)) )= 5/2 -1/4sum_(n=0)^4 (1/2)^n#

This last expression allows us also to calculate the sum, as we have the partial sum of a geometric series:

#sum_(k=0)^n x^k = (1-x^(k+1))/(1-x)#

#S = 5/2 -1/4sum_(n=0)^4 1/2^n = 5/2 - 1/4(1-1/2^5)/(1-1/2) = 5/2- 1/4(31/32)/(1/2) = 5/2- 31/64 = 129/64#