Question

prove by mathematical induction 1³+2³+3³+...n²=(1+2+3+...+n) ² for all n≥1

Answer 1

Let  $n=1$.

$1^3=\mathbf{1} \\ \\ (1)^2=\mathbf{1}$

Both are equal.

Let  $n=2$.

$1^3+2^3=1+8=\mathbf{9} \\ \\ (1+2)^2=3^2=\mathbf{9}$

Both are equal.

Let  $n=3$.

$1^3+2^3+3^3=1+8+27=\mathbf{36}\\ \\ (1+2+3)^2=6^2=\mathbf{36}$

Both are equal.

Let  $n = k$.

Assume that  $1^3+2^3+3^3+...+k^3=(1+2+3+...+k)^2$

Let  $n=k+1$.

$\begin{aligned}&1^3+2^3+3^3+...+n^3\\ \\ \Longrightarrow\ \ &1^3+2^3+3^3+...+(k+1)^3\\ \\ \Longrightarrow\ \ &1^3+2^3+3^3+...+k^3+(k+1)^3\\ \\ \Longrightarrow\ \ &(1+2+3+...+k)^2+(k+1)^3\\ \\ \Longrightarrow\ \ &\left(\frac{k(k+1)}{2}\right)^2+(k+1)(k+1)^2\\ \\ \Longrightarrow\ \ &(k+1)^2\left(\left(\frac{k}{2}\right)^2+k+1\right)\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &(k+1)^2\left(\frac{k^2}{4}+k+1\right)\\ \\ \Longrightarrow\ \ &(k+1)^2\left(\frac{k^2+4k+4}{4}\right)\\ \\ \Longrightarrow\ \ &(k+1)^2\left(\frac{k+2}{2}\right)^2\\ \\ \Longrightarrow\ \ &\left(\frac{(k+1)(k+2)}{2}\right)^2\\ \\ \Longrightarrow\ \ &\left(\frac{(k+1)(k+1+1)}{2}\right)^2\\ \\ \Longrightarrow\ \ &\left(\frac{n(n+1)}{2}\right)^2\\ \\ \Longrightarrow\ \ &(1+2+3+...+n)^2\end{aligned}$

Hence Proved!

Answer 2

$\large{ \sf{ \underline{Answer \colon}}}$

LHS = 1³ = 1

RHS = [ 1² (1 + 1)² ] / 4 = 4/4 = 1

which is true

suppose that n = k is true

then

1³ + 2³ + 3³ + ... + k³ = [ k² (k + 1)² ] /4

now suppose that n = k + 1 is true

LHS

= 1³ + 2³ + 3³ + ... + k³ + (k + 1)³

=[ k² (k + 1)² ] /4 + (k + 1)³

= [ k² (k + 1)² + 4(k + 1)³ ] /4

= [ k² (k + 1)² + 4(k + 1)³ ] /4

= [ k⁴ + 2k³ + k² + 4k³ + 12k² + 12k + 4 ] /4

= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4

RHS

= [ (k + 1)² (k + 2)² ] /4

= [ (k² + 2k + 1)(k² + 4k + 4) ] /4

= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4

LHS = RHS

P(k) = P(k + 1)

so by mathematical induction, the statement is true.