First slide

Introduction to P.M.I

Question

The sum of series 1+2+3+………n is less than

Moderate

A

$\frac{1}{8} (2 n + 1)^{2}$
B

$\frac{1}{12} (2 n + 1)^{2}$
C

$\frac{1}{8} (2 n + 1)$
D

None of these

Solution

Let the given statement be P(n).
$P (n) : 1 + 2 + 3 + \dots + n < \frac{1}{8} (2 n + 1)^{2}$
Step I: For n =1,
$1 < \frac{1}{8} (2 \cdot 1 + 1)^{2} \Rightarrow 1 < \frac{1}{8} \times 3^{2} \Rightarrow 1 < \frac{9}{8}$ which is true.
Step ll :Let it is true for n = k,
$1 + 2 + 3 + \dots + k < \frac{1}{8} (2 k + 1)^{2} - - - (i)$
Step lll: For n =k+1,
$(1 + 2 + 3 + \dots + k) + (k + 1) < \frac{1}{8} (2 k + 1)^{2} + (k + 1)$ [using Eq. (i)]
$\begin{array}{l} = \frac{(2 k + 1)^{2}}{8} + \frac{k + 1}{1} = \frac{(2 k + 1)^{2} + 8 k + 8}{8} \\ = \frac{4 k^{2} + 1 + 4 k + 8 k + 8}{8} = \frac{4 k^{2} + 12 k + 9}{8} \\ = \frac{(2 k + 3)^{2}}{8} = \frac{(2 k + 2 + 1)^{2}}{8} = \frac{[2 (k + 1) + 1]^{2}}{8} \end{array}$
$\Rightarrow 1 + 2 + 3 + \dots + k + (k + 1) < \frac{[2 (k + 1) + 1]^{2}}{8}$
Therefore, P(k + 1) is true when P(k) is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

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