Write the first three terms in a sequence whose $n^{t h}$ term is given by
$c_{n} = \frac{n (n + 1) (2 n + 1)}{6}, \forall n ϵ N$

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Solution

Here, $c_{n} = \frac{n (n + 1) (2 n + 1)}{6}, \forall n ϵ N$
For $n = 1, c_{1} = \frac{1 (1 + 1) (2 (1) + 1)}{6} = 1$ .
For $n = 2, c_{2} = \frac{2 (2 + 1) (4 + 1)}{6} = \frac{2 (3) (5)}{6} = 5$ .
Finally $n = 3, c_{3} = \frac{3 (3 + 1) (7)}{6} = \frac{(3) (4) (7)}{6} = 14$ .
Hence, the first three terms of the sequence are $1, 5$ , and $14$ .
In the above example, we were given a formula for the general term and were able to find any particular term directly. In the following example, we shall see another way of generating a sequence.

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