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Sum of n Terms

Prove: 1+2+...

Question

Prove: $1 + 2 + 3 + 4 + 5 \cdot . . + n = \frac{n (n + 1)}{2}$

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Solution

$1 + 2 + 3 + 4 + 5 + 6 + . . . . + n$

The series is in AP

$\Rightarrow S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
$\Rightarrow \frac{n}{2} (2 + n - 1)$

$\Rightarrow \frac{n (n + 1)}{2}$

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1 + 2 + 3 + 4 + 5 + \dots + n = \frac{n (n + 1)}{2}

n \geq 1

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

1 + \frac{2 n}{3} + \frac{2 n (2 n + 2)}{3.6} + \frac{2 n (2 n + 2) (2 n + 4)}{3.6.9} + . . . . .

= 2^{n} {1 + \frac{n}{3} + \frac{n (n + 1)}{3.6} + \frac{n (n + 1) (n + 2)}{3.6.9} + . . . .}

1 + 2 + 3 + . . . . . + n =

\frac{n (n + 1)}{2}

Q. Using Binomial theorem, prove the inequality

2 \leq {(1 + \frac{1}{n})}^{n} < 3, n \geq 1

n^{n + 1} > (n + 1)^{n}, n \geq 3, n ϵ N

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