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In the proof of the Fundamental Theorem of Calculus we encounter a telescoping sum. Find the values of the following sums, which are also telescoping.

a∑k=11001k-1k+1(b)∑k=1100k2-k-122

Short Answer

Expert verified

Part (a). The required sum is 100101.

Part (b). The required sum is 5000.

Step by step solution

01

Part (a) Step 1. Given Information

We are given the expressions

a∑k=11001k-1k+1(b)∑k=1100k2-k-122

and we need to find their sum.

02

Part (a) Step 2. Explanation

The telescoping sum is:
∑k=11001k-1k+1=-1100+1+11=-1+101101=100101

03

Part (b) Step 1. Explanation

The telescoping sum is:
∑k=1100k2-k-122=1-122+10022=100002=5000

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