Question

Prove by mathematical induction that for every natural number n, 1^3+2^3+3^3+......+n^3=[n(n+1)/2]^2​

Answer 1

Answer:

$\sf\\Solution:\\\textsf{Here, let}\\P(n)=[\frac{n(n+1)}{2}]^2=1^3+2^3+3^3+...+n^3\\\\\textsf{Step 1: First, we check the validity of the given statement P(n) for n = 1.}\\P(1)=[\frac{1(1+1)}{2}]^2=[\frac{2}{2}]^2=1\\\textsf{So, P(1) is true.}$

$\sf\\\textsf{Step 2: Now suppose that the statement P(n) is true for n = k.}\\or,\ P(k)=[\frac{k(k+1)}{2}]^2\\\\\therefore\ 1^3+2^3+3^3+...+k^3=[\frac{k(k+1)}{2}]^2.......(1)$

$\sf\\\textsf{Step 3: Now let us try to establish that P(k+1) is also true.}\\\textsf{For this, add }(k+1)^3\textsf{ on both sides in equation(1).}\\1^3+2^3+3^3+...+k^3+(k+1)^3=[\frac{k(k+1)}{2}]^2+(k+1)^3\\or,\ P(k+1)=(k+1)^2[\frac{k^2}{4}+(k+1)]\\or,\ P(k+1)=(k+1)^2[\frac{k^2+4k+4}{4}]\\or,\ P(k+1)=(k+1)^2[\frac{k+2}{2}]^2\\or,\ P(k+1)=[\frac{(k+1)(k+2)}{2}]^2$

Answer 2

Answer:

Overall steps:

• Verify the base case where $n = 1$.
• Verify the induction step: assuming that $1^{3} + \cdots + k^{3} = (k\, (k + 1) / 2)^{2}$, show that $1^{3} + \cdots + k^{3} + (k + 1)^{3} = ((k + 1)\, (k + 2) / 2)^{2}$.

Step-by-step explanation:

Assuming that $1$ the smallest of all the natural numbers. The base case is to verify that the proposed relation holds for $n = 1$:

$\displaystyle (\text{LHS}) = 1^{3}$.

$\displaystyle (\text{RHS}) = \left(\frac{1\, (1 + 1)}{2}\right)^{2} = 1$.

Hence, the base case is valid.

For the induction step, assume that for some natural number $k$, the proposed relation holds for $n = k$. This assumption is the induction hypothesis:

$\displaystyle 1^{3} + \cdots + k^{3} = \left(\frac{k\, (k + 1)}{2}\right)^{2}$.

The goal is to verify that given the induction hypothesis for $n = k$, the same relation holds for $n = (k + 1)$:

$(\text{LHS}) = 1^{3} + \cdots + k^{3} + (k + 1)^{3}$.

$\displaystyle (\text{RHS}) = \left(\frac{(k + 1)\, ((k + 1) + 1)}{2}\right)^{2} = \left(\frac{(k + 1)\, (k + 2)}{2}\right)^{2}$.

Start by substituting the induction hypothesis for $n = k$ into the $(\text{LHS})$ expression for $n = (k + 1)$:

$\begin{aligned} (\text{LHS}) &= 1^{3} + \cdots + k^{3} + (k + 1)^{3} \\ &= \left(\frac{k\, (k + 1)}{2}\right)^{2} + (k + 1)^{3} && (\text{induction hypothesis}) \\ &= \frac{k^{2}\, (k + 1)^{2}}{2^{2}} + (k + 1)^{2} \, (k + 1) \\ &= (k + 1)^{2}\, \left[\frac{k^{2}}{2^{2}} + (k + 1)\right] \\ &= (k + 1)^{2}\, \left[\frac{k^{2} + 2^{2}\, (k + 1)}{2^{2}}\right] \\ &= (k + 1)^{2} \, \left[\frac{k^{2} + 2\, (2\, k) + 2^{2}}{2^{2}}\right] \\ &= (k + 1)^{2}\, \left[\frac{(k + 2)^{2}}{2^{2}}\right]\end{aligned}$.

Simplify to obtain:

$\displaystyle \left(\frac{(k + 1)\, (k + 2)}{2}\right)^{2}$.

In other words, given that the proposed relationship holds for $n = k$ for any natural number $k$, the same relationship will hold for $n = (k + 1)$.

Hence, by the principle of mathematical induction, the proposed relationship holds for all natural numbers.