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Sigma Notation

Key Questions

  • First expand the series for each value of n #n/(2n+1)#=#1/(2(1)+1#+#2/(2(2)+1#+#3/(2(3)+1#+#4/(2(4)+1#+#5/(2(5)+1#

    Next, perform the operations in the denominator...

    #1/(3)#+#2/5#+#3/7#+#4/9#+#5/11#

    Now, to add fractions we need a common denominator... in this case it's #3465#

    Next, we have to multiply each numerator and denominator by the missing components...

    #1/3# gets multiplied by #1155# giving #1155/3465#

    (Divide the #3465# by #3# to get #1155# and divide the rest by the given denominator.)

    #2/5*693/693=1386/3465, 3/7*495/495=1485/3465, 4/9*385/385=1540/3465 and 5/11*315/315=1575/3465#

    Now simply add the numerators together... #(1155+1386+1485+1540+1575)/3465#

    giving #7141/3465#.

  • #1/2+1/4+1/8+cdots=1/2^1+1/2^2+1/2^3+cdots=sum_{n=1}^infty1/2^n#

  • Sigma notation can be a bit daunting, but it's actually rather straightforward. The common way to write sigma notation is as follows:

    #sum_(x)^nf(x)#

    Breaking it down into its parts:

    • The #sum# sign just means "the sum".
    • The #x# at the bottom is our starting value for x. It usually has a number next to it: #sum_(x=0)#, for example, means we start at x=0 and carry on upwards until we hit...
    • The #n# at the top.
    • The #f(x)# is what we need to plug all these values into. At the end, we add the results obtained from here together, and that's our answer.

    Note that it's not always #f(x)# - it is most often #f(n)# or #f(i)#.

    As an example:

    #sum_(x=0)^9(sqrt(x)+1)^2#

    means we need to find

    #(sqrt(0)+1)^2+(sqrt(1)+1)^2+(sqrt(2)+1)^2+...+(sqrt(9)+1)^2#.

Questions