Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 1C / a

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- Solution a

Solution

Subtract successive terms and use a telescoping series:

z_{n + 1} - z_{1} = \sum_{j = 1}^{n} (z_{j + 1} - z_{j}) = \sum_{j = 1}^{n} c_{j} .

Thus the linear recurrence relation has solution

z_{n + 1} = z_{1} + j = 1 \sum n c_{j} .

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