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Answer 1

Mathematical Induction

We are given statement to prove by the Principle of Mathematical Induction.

The $n^{th}$ term isn't given. We must figure it out.

Each term has sum of cubes in numerator and sum of odd numbers in the denominator.

Observing the pattern, the $n^{th}$ term must contain sum of cubes of n terms in the numerator and sum of first n odd numbers in the denominator.

While proving, we would be using two results:

$\boxed{\begin{minipage}{65 mm} \sf \displaystyle \bullet\quad 1^3+2^3+3^3+\dots+n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\\\\\ \bullet \quad 1+3+5+\dots+(2n-1)=n^2 \end{minipage}}$

The second result above is a simple A.P. with first time 1, common difference 2, and total n terms. So sum can be found easily and it is obtained as $\sf n^2$.

Let us name the statement as P(n). So, we have to prove:

$\sf\displaystyle P(n): \frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots+\frac{1^3+2^3+3^3+\dots n^3}{1+3+5+\dots+(2n-1)} = \frac{n}{24}\left[2n^2+9n+13\right]$

Checking for P(1):

$\begin{array}{l|lc} \mathbb{LHS} & &\mathbb{RHS} \\ \cline{1-3} & \\ \sf = \dfrac{1^3}{1} & \sf = & \sf\dfrac{1}{24} \left[ 2(1)^2+9(1)+13\right] \\ \\ = \sf 1 & \sf = & \sf \dfrac{1}{24}\times 24\\ \\ \sf = 1 & = & \sf 1\end{array}$

Hence, P(1) is true.

Suppose P(k) is true. Then:

$\sf\displaystyle P(k): \frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots+\frac{1^3+2^3+3^3+\dots k^3}{1+3+5+\dots+(2k-1)} = \frac{k}{24}\left[2k^2+9k+13\right] \quad \textsf{---(1)}$

We now need to prove for P(k+1).

We now need to prove for P(k+1). To Prove:

$\sf\displaystyle P(k+1): \frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots+\frac{1^3+2^3+3^3+\dots (k+1)^3}{1+3+5+\dots+(2(k+1)-1)} = \frac{n}{24}\left[2(k+1)^2+9(k+1)+13\right]$

Consider the LHS:

$\mathbb{LHS}\\\\\\\sf\displaystyle = \underbrace{\sf\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots+\frac{1^3+2^3+\dots k^3}{1+3+\dots+(2k-1)}}_{ \textsf{Use (1)}}+\overbrace{\underbrace{\sf\frac{1^3+2^3+\dots+(k+1)^3}{1+3+\dots+(2(k+1)-1)}}_{\sf\textsf{Sum = }(k+1)^2}}^{\sf \textsf{Sum = }\left(\dfrac{(k+1)(k+2)}{2}\right)^2}\\\\\\=\frac{k}{24}\left[2k^2+9k+13\right] + \frac{(k+2)^2\cancel{(k+1)^2}}{4\cancel{(k+1)^2}}\\\\\\=\frac{k}{24}\left[2k^2+9k+13\right] +\frac{(k+2)^2}{4}$

$\sf\displaystyle = \frac{k}{24}\left[2k^2+9k+13\right] +\frac{(k+2)^2}{4}\\\\\\=\frac{1}{4}\left(\frac{2k^3+9k^2+13k}{6}+k^2+4k+4\right)\\\\\\ = \frac{1}{4}\left(\frac{2k^3+9k^2+13k+6k^2+24k+24}{6}\right)\\\\\\=\frac{1}{24}(2k^3+15k^2+37k+24)\\\\\\\textsf{We notice that (k+1) is a factor}\\\\\\=\frac{1}{24}(2k^3+2k^2+13k^2+13k+24k+24)\\\\\\=\frac{1}{24}(2k^2(k+1)+13k(k+1)+24(k+1))\\\\\\=\frac{(k+1)}{24}\left[2k^2+13k+24\right] \\\\\\ =\frac{(k+1)}{24} \left[ 2k^2+4k+2k + 9k + 9 + 13\right]$

$\sf\displaystyle =\frac{(k+1)}{24}\left[2(k^2+2k+1)+9(k+1)+13 \right]\\\\\\ = \frac{(k+1)}{24} \Big[ 2(k+1)^2+9(k+1)+13\Big] \\\\\\ = \mathbb{RHS}$

Thus, P(k+1) is true provided P(k) is true.

Now, P(1) is true and P(k) is true $\implies$ P(k+1) is true.

So, P(n) is true for all $\sf n \in \mathbb{N}$

$\mathcal{HENCE\ \ PROVED}\textsf{, by Principle of Mathematical Induction}$