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So I am given a sequence where the terms $T(n)$ are: $1, 4, 11, 26, 57, 120$ and so on.

Each term is generated by the sum $\sum\limits_{n=1}^k(2^n-1)$

I am being asked to express this sum above in terms of $n$ without the sigma notation, such that I can generate any term $T(n)$ by plugging in a value of $n$ where $n$ belongs to the integers.

I've been stuck in this for a while and I don't know what to do! I would really appreciate some help and I'm slightly short on time. Thanks for the help!

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  • $\begingroup$ What you're being asked for is known as a closed form for the sum. $\endgroup$
    – Brian M. Scott
    Commented May 12, 2012 at 18:02
  • $\begingroup$ As stated by Ross, this decomposes to a geometric sum and an easy sum when you employ the property: $$\sum_{i=1}^{n}a_i+b_i=\sum_{i=1}^{n}a_i+\sum_{i=1}^{n}b_i$$ In general, you may find it useful to look here. $\endgroup$
    – 000
    Commented May 12, 2012 at 18:47

1 Answer 1

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Hint: If you split it $$\sum_{n=1}^k2^n-1=\sum_{n=1}^k2^n-\sum_{n=1}^k1$$ you have one geometric series and one (simple) arithmetic series.

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