The sum of the first n term of a series is #3-[1/3^(n-1)]#. How to obtain the expression for the #n#th term of the series, Un?

Question

Elias Ackerman · Accepted Answer

# U_n=2/3^(n-1), or, 2*3^(1-n)#.

Let, #S_n# denote the sum of the first #n# terms of the seq. #{U_n}#. Then, #S_n=[U_1+U_2+...+U_(n-1)]+U_n#, # rArr S_n=S_(n-1)+U_n#. #:. U_n=S_n-S_(n-1)#, #=[3-1/3^(n-1)]-[3-1/3^((n-1)-1)]#, #=1/3^(n-2)-1/3^(n-1)#, #=1/(3^n/3^2)-1/(3^n/3)#, #=3^2/3^n-3/3^n=(9-3)/3^n#, #=(2xx3)/3^n#, # rArr U_n=2/3^(n-1), or, 2*3^(1-n)#.

Sadie Ackerman · Answer

undefined To obtain the expression for the nth term of the series $ U_n $, subtract the sum of the first $ (n-1) $ terms from the sum of the first $ n $ terms, which gives:

$$ U_n = \text{Sum of first n terms} - \text{Sum of first (n-1) terms} $$

Substituting the given expression for the sum of the first n terms:

$$ U_n = \left(3 - \frac{1}{3^{(n-1)}}\right) - \left(3 - \frac{1}{3^{(n-1) - 1}}\right) $$

$$ U_n = \left(3 - \frac{1}{3^{(n-1)}}\right) - \left(3 - 3^{(n-1)}\right) $$

$$ U_n = \frac{1}{3^{(n-1)}} - 3^{(n-1)} $$

Therefore, the expression for the nth term of the series is:

$$ U_n = \frac{1}{3^{(n-1)}} - 3^{(n-1)} $$