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Request help in understanding the telescoping sum for the given series.
For $n\geq1$, \begin{align} (a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}&=\sum_{i=0}^{n-1}a^{i+1}b^{n-1-i}-\sum_{i=0}^{n-1}a^ib^{n-i}\\ &=\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\\ &=a^n-b^n&&(\text{telescoping sum}) \end{align}

I mean the conversion from the second last step to the last step is not clear.

jiten
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  • I don't mean to offend, but do you understand why telescoping series are called telescoping? (It's a sincere question; lots of people don't get it.) – Brian Tung Jun 16 '19 at 04:05
  • @BrianTung Yes, only the first and last pieces of the telescope remain. But, very difficult to see here how it happens. – jiten Jun 16 '19 at 04:07
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    I find that it's usually effective to pick an $n$ and explicitly write out some terms. – Brian Tung Jun 16 '19 at 04:09

2 Answers2

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Given$$ \sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\\$$ Change the index of summation on the first sum: $$=\sum_{i=1}^na^ib^{n-i} - \sum_{i=0}^{n-1}a^ib^{n-i}$$ Take out the $i=n$ term from the first sum and the $i=0$ term from the second, combining the rest:

$$=a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n$$

$$=a^n+0-b^n = a^n-b^n$$

J. W. Tanner
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  • Thanks, but my post had origins in solving the below problem that I hoped to solve using telescoping. But, am unable till now. Please tell if the below one can be solved using telescoping technique or not. If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using your approach. I tried as follows, but was unsuccessful: $a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$ – jiten Jun 16 '19 at 07:19
  • You're welcome. It appears that you have now (appropriately) posted that separate question – J. W. Tanner Jun 17 '19 at 02:29
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If you write $c_i = a^ib^{n-i}$, then your sum turns into $$\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i}) = \sum_{i=0}^{n-1}(c_{i+1}-c_i) = c_n - c_0 = a^n-b^n$$

trancelocation
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