Math, asked by deepaktsatna, 9 months ago



 \frac{3}{ {1}^{2}  \times  {2}^{2} }  \times  \frac{5}{ {2}^{2} \times  {3}^{2}  }  \times  \frac{7}{ {3}^{2} \times  {4}^{2}  }  \times  \frac{19}{ {9}^{2}  \times  {10}^{2} }

Answers

Answered by ManuAgrawal01
87

Given:-

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  \bf \implies\frac{3}{ {1}^{2} \times {2}^{2} } \times \frac{5}{ {2}^{2} \times {3}^{2} } \times \frac{7}{ {3}^{2} \times {4}^{2} } \times \frac{19}{ {9}^{2} \times {10}^{2} }

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To Find:-

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 \bf \implies The  \: Solution

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STEP BY STEP EXPLANATION:-

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  \bf \implies\frac{3}{ {1} \times {2}^{2} } \times \frac{5}{ {2}^{2} \times {3}^{2} } \times \frac{7}{ {3}^{2} \times {4}^{2} } \times \frac{19}{ {9}^{2} \times {10}^{2} }

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  \bf \implies\frac{1}{ {1} \times {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{ {3}^{2} \times {4}^{2} } \times \frac{19}{ {9}^{2} \times {10}^{2} }

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  \bf \implies\frac{1}{ {1} \times {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{ (3 \times 4 {)}^{2} } \times \frac{19}{ {9}^{2} \times {10}^{2} }

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  \bf \implies\frac{1}{ {1} \times {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{ (3 \times 4 {)}^{2} } \times \frac{19}{( {9} \times 10{)}^{2} }

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  \bf \implies\frac{1}{  {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{ (3 \times 4 {)}^{2} } \times \frac{19}{( {9} \times 10{)}^{2} }

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  \bf \implies\frac{1}{  {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{  {12}^{2} } \times \frac{19}{( {9} \times 10{)}^{2} }

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  \bf \implies\frac{1}{  {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{  {12}^{2} } \times \frac{19}{{90}^{2} }

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  \bf \implies\frac{1}{  {2}^{2} } \times \frac{5}{ {2}^{2} \times {3} } \times \frac{7}{  144} \times \frac{19}{{90}^{2} }

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  \bf \implies \frac{665}{432 \times  {2}^{4}  \times  {90}^{2} }

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  \bf \implies \frac{665}{432 \times  16 \times  {90}^{2} }

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  \bf \implies \frac{665}{6912 \times  {90}^{2} }

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  \bf \implies \frac{665}{6912 \times  8100 }

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  \bf \implies \frac{133}{6912 \times  1620}

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  \bf \implies \frac{133}{11197440}

Answered by Saby123
19

To solve -

Find the required value of -

=> { [ 3 ] / [ 1² × 2² ] } × { [ 5 ] / [ 2² × 3² ] } × { [ 7 ] / [ 3² × 4² ] } × { [ 19 ] / [ 9² + 10² ] } .

Solution -

Here we have to find the value of -

=> { [ 3 ] / [ 1² × 2² ] } × { [ 5 ] / [ 2² × 3² ] } × { [ 7 ] / [ 3² × 4² ] } × { [ 19 ] / [ 9² + 10² ] } .

We can observe that any term of the given sequence can be represented as - .

=> [ a + b ] / [ a² + b² ]

But here as the terms are not consecutive , we have to solve manually.

=> { [ 3 ] / [ 1² × 2² ] } × { [ 5 ] / [ 2² × 3² ] } × { [ 7 ] / [ 3² × 4² ] } × { [ 19 ] / [ 9² + 10² ] } .

=> { [ 1 ] / [ 1² × 2² ] } × { [ 5 ] / [ 2² × 3 ] } × { [ 7 ] / [ 3² × 4² ] } × { [ 19 ] / [ 9² + 10² ] }

=> { [ 1 ] / [ 4 ] } × { [ 5 ] / [ 12 ] } × { [ 7 ] / [ 144] } × { [ 19 ] / [ 8100 ] }

=> [ 665 ] / [ 4 × 12 × 144 × 8100 ]

=> [ 665 ] / [ 11197440 ] .

This is the required answer .

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Additional Information -

( a + b )² = a² + 2ab + b²

( a - b )² = a² - 2ab + b²

( a + b )( a - b ) = a² - b²

( a + b )³ = a³ + 3ab ( a + b ) + b³

( a - b )³ = a³ - 3ab ( a + b ) - b³

( a + b + c )³ = a³ + b³ + c³ + 3 ( a + b )( b + c )( c + a )

a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² - ab - bc - ca )

When a + b + c = 0 ,

a³ + b³ + c³ = 3abc .

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