Prove 2 - 5 - 8 - 11 - - - 3n - 1 - 12 n 3n - 1 n - N

Let P(n) be true for n = m , that is, we suppose that #160; P(m) = 2 + 5 + 8 + 11 + ... + (3m 1) = 12 m (3m + 1)Now P(m + 1) = P(m) + T ...

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Proof by mathematical induction

Prove 2 + 5...

Question

Prove $2 + 5 + 8 + 11 + . . . + (3 n - 1) =$ $\frac{1}{2}$ $n (3 n + 1) n$ $ϵ$ N

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Prove using mathematical induction that for all $n \geq 1$

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

n \in N

\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{(6 n + 4)}

2^{2} \frac{C_{0}}{1.2} + 2^{3} \frac{C_{1}}{2.3} + 2^{4} \frac{C_{2}}{3.4} + . . . . . . . . . + 2^{n + 2} \frac{C_{n}}{(n + 1) (n + 2)} = \frac{3^{n + 2} - 2 n - 5}{(n + 1) (n + 2)}

Prove by using the principle of mathematical induction $\forall n \in N$

$2 + 5 + 8 + 11 + . . . + (3 n - 1) = \frac{1}{2} n (3 n + 1)$

Using principle of mathematical induction, prove that $4^{n} + 15 n - 1$ is divisible by 9 for all natural numbers n.

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