1.

Using binomial theorem, prove that `(2^(3n)-7n-1)` is divisible by 49, where n `in` N.

Answer» `( 1 + x ) ^ ( n ) = . ^ ( n) C _ ( 0 ) + .^ (n) C _ ( 1 ) x + .^ ( n ) C _ ( 2 ) x ^ ( 2 ) + ...+ .^ ( n ) C _ ( n)x ^ ( n ) .`
`( 1 + 7 ) ^ ( n ) = . ^ ( n) C _ ( 0 ) + . ^ ( n) C_ ( 1 ) xx 7 + . ^ ( n ) C _( 2 ) xx ( 7 ) ^ ( 2 ) + ...+ . ^ ( n ) C _ ( n ) xx ( 7 ) ^ ( n ) `
`= 1 + 7 n + ( 7 ) ^ ( 2 ) [.^ ( n) C_ ( 2 ) + .^ ( n) C_ ( 3 ) xx 7 + ... + ( 7 ) ^ ( n-2 ) ]`
`rArr ( 8 ^ ( n ) - 7n _ 1 ) = ( 7 ) ^ ( 2 ) xx " (an integar)" = 49 xx "( an integar) ".` ,
Hence, `(2 ^ ( 3n ) -7n-1)` is divisible by 49.


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