Using mathematical induction, prove...

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1993Moderate

Using mathematical induction, prove that $tan^{- 1} (1/3) + tan^{- 1} (1/7) + .... tan^{- 1} {1/ (n^{2} + n + 1)} = tan^{- 1} {n / (n + 2)}$

Visualized Solution (Hindi)

Defining the Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (n)

Let the given statement be $P (n)$ :
$\sum_{r = 1}^{n} tan^{- 1} \frac{1}{r ^{2} + r + 1} = tan^{- 1} \frac{n}{n + 2}$
Our goal is to prove this for all $n \in N$ using induction.

Base Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n = 1

Base Case: Check for $n = 1$ .
LHS: $tan^{- 1} \frac{1}{1 ^{2} + 1 + 1} = tan^{- 1} \frac{1}{3}$
RHS: $tan^{- 1} \frac{1}{1 + 2} = tan^{- 1} \frac{1}{3}$
Since LHS = RHS, $P (1)$ is true.

Inductive Hypothesis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (k)

Inductive Hypothesis: Assume $P (k)$ is true for some $k \in N$ .
$\sum_{r = 1}^{k} tan^{- 1} \frac{1}{r ^{2} + r + 1} = tan^{- 1} \frac{k}{k + 2}$

Inductive Step Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (k + 1)

Inductive Step: Consider $P (k + 1)$ .
Sum of $k + 1$ terms = (Sum of $k$ terms) + ( $(k + 1)^{t h}$ term)
$P (k + 1) = tan^{- 1} \frac{k}{k + 2} + tan^{- 1} \frac{1}{( k + 1 ) ^{2} + ( k + 1 ) + 1}$

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (k + 1)^{t h} Term

Simplify the denominator of the new term:
$(k + 1)^{2} + (k + 1) + 1 = k^{2} + 2 k + 1 + k + 1 + 1 = k^{2} + 3 k + 3$
So, $P (k + 1) = tan^{- 1} \frac{k}{k + 2} + tan^{- 1} \frac{1}{k ^{2} + 3 k + 3}$

Applying Inverse Tangent Formula

Apply the formula: $tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}$
$P (k + 1) = tan^{- 1} [\frac{\frac{k}{k + 2} + \frac{1}{k ^{2} + 3 k + 3}}{1 - \frac{k}{( k + 2 ) ( k ^{2} + 3 k + 3 )}}]$

Expanding the Numerator

Numerator: $\frac{k ( k ^{2} + 3 k + 3 ) + 1 ( k + 2 )}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$
$= \frac{k ^{3} + 3 k ^{2} + 3 k + k + 2}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$
$= \frac{k ^{3} + 3 k ^{2} + 4 k + 2}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$

Expanding the Denominator

Denominator: $1 - \frac{k}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$
$= \frac{( k + 2 ) ( k ^{2} + 3 k + 3 ) - k}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$
$= \frac{k ^{3} + 3 k ^{2} + 3 k + 2 k ^{2} + 6 k + 6 - k}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$
$= \frac{k ^{3} + 5 k ^{2} + 8 k + 6}{( k + 2 ) ( k ^{2} + 3 k + 3 )}$

Factoring and Final Result

$P (k + 1) = tan^{- 1} [\frac{k ^{3} + 3 k ^{2} + 4 k + 2}{k ^{3} + 5 k ^{2} + 8 k + 6}]$
Factorizing: $k^{3} + 3 k^{2} + 4 k + 2 = (k + 1) (k^{2} + 2 k + 2)$
Factorizing: $k^{3} + 5 k^{2} + 8 k + 6 = (k + 3) (k^{2} + 2 k + 2)$
$P (k + 1) = tan^{- 1} \frac{k + 1}{k + 3} = tan^{- 1} \frac{k + 1}{( k + 1 ) + 2}$

Conclusion

Conclusion: Since $P (1)$ is true and $P (k) ⟹ P (k + 1)$ , the statement is true for all $n \in N$ .
Key Takeaway: The sum of this inverse tangent series follows a predictable rational pattern.
Challenge: Try to prove this using the telescoping series method by writing $\frac{1}{n ^{2} + n + 1}$ as $\frac{( n + 1 ) - n}{1 + n ( n + 1 )}$ .

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Using mathematical induction, prove that $tan^{- 1} (1/3) + tan^{- 1} (1/7) + .... tan^{- 1} {1/ (n^{2} + n + 1)} = tan^{- 1} {n / (n + 2)}$

Visualized Solution (Hindi)

Defining the Statement $P (n)$

Base Case $n = 1$

Inductive Hypothesis $P (k)$

Inductive Step Setup $P (k + 1)$

Simplifying the $(k + 1)^{t h}$ Term

Applying Inverse Tangent Formula

Expanding the Numerator

Expanding the Denominator

Factoring and Final Result

Conclusion

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using mathematical induction, prove that tan−1(1/3)+tan−1(1/7)+....tan−1{1/(n2+n+1)}=tan−1{n/(n+2)}

Visualized Solution (Hindi)

Defining the Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(n)

Base Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=1

Inductive Hypothesis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(k)

Inductive Step Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(k+1)

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (k+1)th Term

Applying Inverse Tangent Formula

Expanding the Numerator

Expanding the Denominator

Factoring and Final Result

Conclusion

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Base Case $n = 1$

Inductive Hypothesis $P (k)$

Inductive Step Setup $P (k + 1)$

Simplifying the $(k + 1)^{t h}$ Term

Prove by mathematical induction that $\frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \leq \frac{1}{( 3 n + 1 ) ^{1/2}}$ for all positive integers $n$ .

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

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