Animated Solution for Mathematics - Inverse Trigonometric Functions: Using mathematical induction, prove that ${\tan }^{-1}(1/3)+{\tan }^{-1}(1/7)+....{\tan }^{-1}\{1/(n^2+n+1)\}={\tan }^{-1}\{n/(n+2)\}$

The series ${\sum }_{r=1}^n{\tan }^{-1}\frac{1}{r^2+r+1}$ appears complex, but it possesses a rhythmic, predictable structure. We aim to prove that this sum is exactly ${\tan }^{-1}\frac{n}{n+2}$ using the principle of mathematical induction.

Every journey begins with a single step. For $n=1$, our series consists of only the first term:

On the right-hand side, substituting $n=1$ into the formula ${\tan }^{-1}\frac{n}{n+2}$ yields:

The base case holds! This confirms that our hypothesis is grounded in reality.

We assume the statement $P(k)$ is true for some arbitrary natural number $k$. This is our Inductive Hypothesis:

We now examine $P(k+1)$. We must show that the sum of the first $k+1$ terms equals ${\tan }^{-1}\frac{k+1}{(k+1)+2}$.

The sum of $k+1$ terms is the sum of the first $k$ terms plus the $(k+1{)}^{th}$ term. Substituting our hypothesis, we obtain:

Let us simplify the $(k+1{)}^{th}$ term. Expanding the denominator, $(k+1{)}^2+(k+1)+1$ becomes $k^2+2k+1+k+1+1$, which simplifies to $k^2+3k+3$.

We now face the addition of two inverse tangents:

We invoke the identity ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$. The numerator becomes:

The denominator becomes:

When we divide the numerator by the denominator, the common term $(k+2)(k^2+3k+3)$ cancels out. This leaves us with:

Both polynomials share the factor $(k^2+2k+2)$. Factoring them out, we get:

This is exactly ${\tan }^{-1}\frac{k+1}{(k+1)+2}$, which is our target $P(k+1)$. By the principle of mathematical induction, the statement is true for all $n\in \mathbb{N}$.