Mathematical Induction

2010-04-03 11:42 pm
Prove by Mathematical Induction
" 5ⁿ(4n-1)+1 is divisible by 16. "

回答 (1)

2010-04-04 12:06 am
✔ 最佳答案
When n = 1 ,
(5^1)(4(1)-1)+1 = 16 is divisible by 16 is true. (Let it be 16T)
Let (5^k)(4k-1)+1 be true , when n = k+1 :
( 5^(k+1) )(4(k+1) - 1) + 1
= 5 (5^k) (4k - 1 + 4) + 1
= 5 (5^k) (4k - 1) + 20(5^k) + 1
= 5 (5^k) (4k - 1) + 5 + 20(5^k) - 4
= 5 [ (5^k) (4k - 1) + 1 ] + 4(5(5^k) - 1)
= 5 [ 16T ] + 4(5(5^k) - 1)
= 5 [ 16T ] + 4(5^(k+1) - 1)
= 5 [ 16T ] + 4((4+1)^(k+1) - 1)
= 5 [ 16T ] + 4[ 4^(k+1) + (k+1)C1 (4^k) + (k+1)C2 (4^k-1) + ... + 1 - 1]
= 5 [ 16T ] + 4{ 4S } where S = 4^k + (k+1)C1 (4^(k-1)) + (k+1)C2 (4^k-2)+..
= 5 [ 16T ] + 16S
= 16 (5T + S) is divisible by 16.


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