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Algebra Level 4

The number of integer terms in the expansion of ( 3 + 5 1 8 ) 256 \left( \sqrt{3}+5^{\frac{1}{8}} \right)^{256} is?

33 34 32 35

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Mudit Bansal
Feb 25, 2015

Consider, T r + 1 = C r 256 3 256 − r 2 5 r 8 { T }_{ r+1 }={ C }_{ r }^{ 256 }{ 3 }^{ \frac { 256-r }{ 2 } }{ 5 }^{ \frac { r }{ 8 } } Now for integral values: 256 − r = 2 I ( I ∈ i n t e g e r ) r = 0 , 2 , 4 , 6 , . . . . ( 1 ) a n d r 8 = k ( k ∈ i n t e g e r ) r = 0 , 8 , 16 , . . . . ( 2 ) 256-r=2I(I\in integer)\\ r=0,2,4,6,....\left( 1 \right) \\ and\quad \frac { r }{ 8 } =k(k\in integer)\\ r=0,8,16,....\left( 2 \right) Now ( 1 ) ∩ ( 2 ) g i v e s \left( 1 \right) \cap \left( 2 \right) \quad gives : r = 0 , 8 , 16 , . . . . 256 r=0,8,16,....256 Hence total number of integral terms is: = 256 8 + 1 = 33 =\frac { 256 }{ 8 } +1=\boxed { 33 }

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Done in the same way, nice solution!!

Bhargav Upadhyay - 6 years, 3 months ago

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