Prove that n⁷/7 + n⁵/5 + 2n³/3 - n/105...

MathematicsBinomial Expansion for Positive Integral IndexJEE Advanced 1990Moderate

Prove that $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105}$ is an integer for every positive integer $n$ .

Visualized Solution (Hindi)

Prove .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105} \in Z

Expression: $E = \frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105}$
Goal: Prove $E \in Z$ for all $n \in Z^{+}$

Common Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 105

$P (n) = \frac{15 n ^{7} + 21 n ^{5} + 70 n ^{3} - n}{105}$

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

For $n = 1$ :
Numerator $= 15 (1)^{7} + 21 (1)^{5} + 70 (1)^{3} - 1 = 105$
Result: $P (1) = \frac{105}{105} = 1 \in Z$

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

Assume $P (k) = \frac{15 k ^{7} + 21 k ^{5} + 70 k ^{3} - k}{105}$ is an integer.

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

To prove $P (k + 1) \in Z$ , we examine the difference:
Difference $D = P (k + 1) - P (k)$

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

$D = \frac{15 (( k + 1 ) ^{7} - k ^{7} ) + 21 (( k + 1 ) ^{5} - k ^{5} ) + 70 (( k + 1 ) ^{3} - k ^{3} ) - (( k + 1 ) - k )}{105}$

Binomial Expansion of Terms

$(k + 1)^{7} - k^{7} = 7 k^{6} + 21 k^{5} + 35 k^{4} + 35 k^{3} + 21 k^{2} + 7 k + 1$
$(k + 1)^{5} - k^{5} = 5 k^{4} + 10 k^{3} + 10 k^{2} + 5 k + 1$
$(k + 1)^{3} - k^{3} = 3 k^{2} + 3 k + 1$

Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 105

$15 (7 k^{6} + \dots + 1) = 105 k^{6} + \dots + 15$
$21 (5 k^{4} + \dots + 1) = 105 k^{4} + \dots + 21$
$70 (3 k^{2} + 3 k + 1) = 210 k^{2} + 210 k + 70$
Constant sum: $15 + 21 + 70 - 1 = 105$

Final Conclusion

Numerator of $D$ is a multiple of $105 ⟹ D \in Z$
$∵ P (k + 1) = P (k) + D$ and $P (k), D \in Z$
$∴ P (k + 1) \in Z$
Key Takeaway: Induction and Binomial Theorem combined are powerful for divisibility proofs.
Next Challenge: Try proving this using Fermat's Little Theorem: $n^{p} \equiv n (mod p)$ .

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Prove that $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105}$ is an integer for every positive integer $n$ .

Visualized Solution (Hindi)

Prove $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105} \in Z$

Common Denominator $105$

Base Case: $n = 1$

Inductive Hypothesis: $n = k$

Inductive Step: $n = k + 1$

Difference $D = P (k + 1) - P (k)$

Binomial Expansion of Terms

Multiples of $105$

Final Conclusion

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Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that 7n7​+5n5​+32n3​−105n​ is an integer for every positive integer n.

Visualized Solution (Hindi)

Prove .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7n7​+5n5​+32n3​−105n​∈Z

Common Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 105

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; } n=k

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

Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=P(k+1)−P(k)

Binomial Expansion of Terms

Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 105

Final Conclusion

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Prove that $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105}$ is an integer for every positive integer $n$ .

Prove $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105} \in Z$

Common Denominator $105$

Base Case: $n = 1$

Inductive Hypothesis: $n = k$

Inductive Step: $n = k + 1$

Difference $D = P (k + 1) - P (k)$

Multiples of $105$

Prove by permutation or otherwise $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ is an integer ( $n \in I^{+}$ ).

Prove that $7^{2 n} + 2^{3 n - 3} (3^{n - 1})$ is divisible by 25 for any natural number $n$ .

Prove that $\sum_{k = 1}^{n - 1} (n - k) cos \frac{2 k π}{n} = - n /2$ , where $n \geq 3$ is an integer.

If $p$ be a natural number then prove that $p^{n + 1} + (p + 1)^{2 n - 1}$ is divisible by $p^{2} + p + 1$ for every positive integer $n$ .

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 $n$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

Prove that for any positive integer $k$ , $\frac{s i n 2 k x}{s i n x} = 2 [cos x + cos 3 x + \dots + cos (2 k - 1) x]$ . Hence prove that $\int_{0}^{π /2} sin 2 k x cot x d x = \frac{π}{2}$

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Prove that for any positive integer $k$ , $\frac{s i n 2 k x}{s i n x} = 2 [cos x + cos 3 x + \dots + cos (2 k - 1) x]$ . Hence prove that $\int_{0}^{π /2} sin 2 k x cot x d x = \frac{π}{2}$