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?

Answer 1

Elias Ackerman

# 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)#.

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Answer 2

Sadie Ackerman

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:

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.