First slide

Introduction to P.M.I

Question

The identity $1^{3} + 2^{3} + 3^{3} + \dots + n^{3}$ is equal to

Easy

A

${[\frac{n (n - 1)}{2}]}^{2}$
B

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

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

${[\frac{n (n + 1)}{2}]}^{2}$

Solution

Let the given statement be P(n).
$P (n) : 1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = {[\frac{n (n + 1)}{2}]}^{2}$
Step I : For n=1,
$P (1) : {[\frac{1 (1 + 1)}{2}]}^{2} = {[\frac{1 \times 2}{2}]}^{2} = 1^{2} = 1 = 1^{3}$ which is true
Step ll: Let it is true for n = k,
$1^{3} + 2^{3} + 3^{3} + \dots + k^{3} = {[\frac{k (k + 1)}{2}]}^{2} - - - - - - (i)$
Step lll: For n=k+1,
$(1^{3} + 2^{3} + 3^{3} + 4^{3} + \dots + k^{3}) + (k + 1)^{3}$
$= {[\frac{k (k + 1)}{2}]}^{2} + (k + 1)^{3}$ [using Eq. (i)]
$\begin{aligned} = \frac{k^{2} (k + 1)^{2}}{4} + \frac{(k + 1)^{3}}{1} \\ = \frac{k^{2} (k + 1)^{2} + 4 (k + 1)^{3}}{4} \end{aligned}$
On taking (k + 1)² common in numerator Part,
$\begin{array}{l} = \frac{(k + 1)^{2} [k^{2} + 4 (k + 1)]}{4} \\ = \frac{(k + 1)^{2} (k^{2} + 4 k + 4)}{4} \\ = \frac{(k + 1)^{2} (k + 2)^{2}}{4} \\ = \frac{(k + 1)^{2} [(k + 1) + 1]^{2}}{4} \\ = {[\frac{(k + 1) {(k + 1) + 1}}{2}]}^{2} \end{array}$
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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