Binomial theorem for positive integral Index

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Question

$If n is a natural number, then 4^{n} - 3 n - 1 is divisible by which one of the following?$

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Solution

$Putting, n = 2, we get 4^{n} - 3 n - 1 = 16 - 6 - 1 = 9$
$Putting, n = 3, we get 4^{n} - 3 n - 1 = 64 - 9 - 1 = 54$
$Putting, n = 4, we get 4^{n} - 3 n - 1 = 256 - 12 - 1 = 243$
$Hence, for any value of n, 4^{n} - 3 n - 1 is$ $always divisible by 9 .$
$s e c o n d m e t h o d : {(1 + 3)}^{n} - 3 n - 1 = [1 + C (n, 1) 3 + C (n, 2) 3^{2} + C (n, 3) 3^{3} + . . . . . + 3^{n}] - 3 n - 1$
= $9 [C (n, 2) + C (n, 3) 3 . . . . . . . .] w h i c h i s d i v i s i b l e b y 9$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If n is a natural number, then 4n-3n-1 is divisible by which one of the following?

Putting, n=2, we get 4n-3n-1=16-6-1=9 Putting, n=3 , we get 4n-3n-1=64-9-1=54 Putting, n=4, we get 4n-3n-1=256-12-1=243 Hence, for any value of n,4n-3n-1 is always divisible by 9 . second method: 1+3n-3n-1= 1+Cn,13+Cn,232+Cn,333+.....+3n-3n-1 = 9Cn,2+Cn,33........ which is divisible by 9

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$If n is a natural number, then 4^{n} - 3 n - 1 is divisible by which one of the following?$