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Evaluate if some formulas for finite sums are true or false

by RoRi
    Decide whether the following are true or false:

  1. \displaystyle{\sum_{n=0}^{100} n^4 = \sum_{n=1}^{100} n^4}.
  2. \displaystyle{\sum_{j=0}^{100} 2 = 200}.
  3. \displaystyle{\sum_{k=0}^{200} (2+k) = 2 + \sum_{k=0}^{100} k}.
  4. \displaystyle{\sum_{i=1}^{100} (i+1)^2 = \sum_{i=0}^{99} i^2}.
  5. \displaystyle{\sum_{k=1}^{100} k^3 = \left(\sum_{k=1}^{100} k \right) \left( \sum_{k=1}^{100} k^2 \right)}.
  6. \displaystyle{\sum_{k=0}^{100} k^3} = \left( \sum_{k=0}^{100} \right)^3}.
  1. True. Since 0^4 = 0 the extra term on the left has no impact on the sum. In other words,

        \[ \sum_{n=0}^{100} n^4 = 0^4 + \sum_{n=1}^{100} n^4 = \sum_{n=1}^{100} n^4. \]

  2. False. We know from here that \sum_{j=1}^{100} 1 = 100; hence,

        \[ \sum_{j=0}^{100} 2 = 2 \left(1 + \sum_{j=1}^{100} 1 \right) = 202. \]

    (The “problem” here is the extra term in the sum. The j=0 term actually contributes to the sum since 2 evaluated at j=0 is still 2.)

  3. False. This is just abusing the additive property of finite sums. In reality,

        \[ \sum_{k=0}^{200} (2+k) = \left( \sum_{k=0}^{200} 2 \right) + \left( \sum_{k=0}^{200} k \right) = 202 + \sum_{k=0}^{100} k. \]

  4. False. This is reindexing the sum incorrectly. Seeing, (i+1)^2 the idea is to replace a sum over i, by a sum over i+1 to get a simpler expression. (A quick and dirty way to see that this has not been reindexed correctly is by looking at the first few terms. The sum on the left starts 2^2 + 3^2 + 4^2 + \cdots, while the sum on the right starts 0^2 + 1^2 + 2^2 + \cdots. To do this correctly, we replace i+1 by i in the interior, and then have to add one to the top and bottom indices so that we have i+1=2 to 101 and can replace the i+1 by i. This is the way I think about it; there are probably better ways.) Anyway, reindexing properly we would get,

        \[ \sum_{i=1}^{100} (i+1)^2 = \sum_{i=2}^{101} i^2. \]

    Or we could actually expand out the (i+1)^2 squared term and use linearity to separate the sums and evaluate all of them separately, but that would be slower.

  5. False. Sums and powers just don’t work this way. (This is really claiming (1^3 + 2^3 + \cdots + 100^3) = (1+2 + \cdots +100)(1^2 + 2^2 + \cdots + 100^2).) From work we did earlier (see: sums of integers, sums of squares, and sums of cubes) we have,

        \[ \sum_{k=1}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=1}^n k \right) \left( \sum_{k=1}^n k^2 \right) = \left( \frac{n^2}{2} + \frac{n}{2} \right) \left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right). \]

    One can then plug in n=100 to see that we get 25502500 on the left, but 1708667500 on the right.

  6. False. Again, this just wrong. The claim is that (1^3 + 2^3 + \cdots + 100^3) = (1 + 2 + \cdots + 100)^3, which is of course false. More concretely, we know

        \[ \sum_{k=0}^{n} k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=0}^n k \right)^3 = \left( \frac{n^2}{2} + \frac{n}{2} \right)^3.\]

    So, evaluating at n=100, we have 25502500 on the right and 128787625000 on the right.

2 comments

  1. Daniel Fugisawa says:

    Hi! On item b. at your answer there seems to be a typo. You wrote the sum of 2 running from 0 to 200. But the problem states the sum of 2 running from 0 to 100. Anyway, that didn’t stopped you from getting to the same answer I did.

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