An algebra problem by hansraj sharma

Algebra Level 5

What is the value of

( 1 + 3 4 + 3 4 × 5 8 + 3 4 × 5 8 × 7 12 + … ) 2 ? \left(1+\frac { 3 }{ 4 } +\frac { 3 }{ 4 } \times \frac { 5 }{ 8 } +\frac { 3 }{ 4 } \times \frac { 5 }{ 8 } \times \frac { 7 }{ 12 } + \ldots\right)^2 ?


The answer is 8.

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Hansraj Sharma by hansraj sharma , 25, India

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1 solution

Hansraj Sharma
Aug 20, 2014

( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 ! x 2 + . . . . . . . ∞ C o m p a r e t h e b a s e o f g i v e n t e r m ∴ n x = 3 4 a n d n ( n − 1 ) 2 x 2 = 3 4 . 5 8 s o l v i n g n = − 3 2 , x = − 1 2 ∴ g i v e n t e r m = [ ( 1 − 1 2 ) − 3 2 ] 2 = ( 1 2 ) − 3 = 2 3 = 8 (1+x{ ) }^{ n }=1+nx+\frac { n(n-1) }{ 2! } { x }^{ 2 }+.......\infty \\ Compare\quad the\quad base\quad of\quad given\quad term\\ \therefore \quad nx=\frac { 3 }{ 4 } and\frac { n(n-1) }{ 2 } { x }^{ 2 }=\frac { 3 }{ 4 } .\frac { 5 }{ 8 } \quad solving\quad n=-\frac { 3 }{ 2 } ,x=-\frac { 1 }{ 2 } \\ \therefore given\quad term\quad =\quad \left[ \left( 1-\frac { 1 }{ 2 } \right) ^{ -\frac { 3 }{ 2 } } \right] ^{ 2 }\quad =\quad \left( \frac { 1 }{ 2 } \right) ^{ -3 }=\quad { 2 }^{ 3 }=\quad 8

1 Helpful 0 Interesting 0 Brilliant 0 Confused

( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 ! x 2 + . . . . . . . ∞ C o m p a r e t h e b a s e o f g i v e n t e r m ∴ n x = 3 4 a n d n ( n − 1 ) 2 x 2 = 3 4 . 5 8 s o l v i n g n = − 3 2 , x = − 1 2 ∴ g i v e n t e r m = [ ( 1 − 1 2 ) − 3 2 ] 2 = ( 1 2 ) − 3 = 2 3 = 8 (1+x{ ) }^{ n }=1+nx+\frac { n(n-1) }{ 2! } { x }^{ 2 }+.......\infty \ Compare\quad the\quad base\quad of\quad given\quad term\ \therefore \quad nx=\frac { 3 }{ 4 } and\frac { n(n-1) }{ 2 } { x }^{ 2 }=\frac { 3 }{ 4 } .\frac { 5 }{ 8 } \quad solving\quad n=-\frac { 3 }{ 2 } ,x=-\frac { 1 }{ 2 } \ \therefore given\quad term\quad =\quad \left[ \left( 1-\frac { 1 }{ 2 } \right) ^{ -\frac { 3 }{ 2 } } \right] ^{ 2 }\quad =\quad \left( \frac { 1 }{ 2 } \right) ^{ -3 }=\quad { 2 }^{ 3 }=\quad 8 @hansraj sharma solution not mine upvote his

Mardokay Mosazghi - 6 years, 9 months ago

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