Prove the following by using the principle of mathematical induction for all n ∈ N:

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Solution

Let the given statement be P(n), i.e.,

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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n \in N : \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}

n \in N :

1 + 3 + 3^{2} + . . . + 3^{n - 1} = \frac{(3^{n} - 1)}{2}

Prove $\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \dots + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}$

\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}

n \in N .

\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}

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