數學歸納法

Question

證明1*3+2*(3^2)+3*(3^3)+...+n*(3^n)=(3/4){1+[2n-1](3^n)}成立
證明(5^n)(4n-1)+1能被16整除
只須列(當n=k+1)的步驟

Answer 1

let it is true for n=k, hence we have 1*3+2*3^2+......k*(3^k) = (3/4){1+ (2k-1)(3^k) )

for n= k+1
L.H.S =1*3+.........k*(3^k) + (k+1)(3^(k+1))
= (3/4){1+(2k-1)(3^k)) + (k+1)(3^(k+1))
=(3/4){ 1+ (2k-1)(3^k) + 4(k+1)(3^k)}
=(3/4)(1+(3^k)(2k-1+4k+4))
=(3/4)(1+(3^k)(6k+3))
=(3/4)(1+(2k+1)(3^(k+1)) =R.H.S
therefore by MI, it is true for all positive no. n


2.

let it is true for n=k, hence we have (5^k)(4k-1)+1 = 16M

for n =k+1
(5^(k+1))(4(k+1)-1) +1

=5( 5^k(4k-1 + 4)) +1
= 5(5^k(4k-1) + 4x5^k)+1
=5(5^k(4k-1)+1 -1+4x5^k) +1
=5(16M -1+4x5^k)+1
=5*16M -5 +4*5^(k+1) +1
=5*16M + 4*5^(k+1) -4
=5*16M + 4(5^(k+1)-1)
= 5*16M+4(5-1)(5^k+5^(k-1)+.....+1)
=5*16M+16(5^k+5^(k-1)+.....+1)
=(16)(5M+5^k+5^(k-1)+.....+1) is divisible by 16

therefore, it is divisible by 16 for all positive no. n

Answer 2

when n=k+1

LHS=1*3+2*(3^2)+3*(3^3)+...+k*(3^k)+(k+1)*[3^(k+1)]
=(3/4)[1+(2k-1)(3^k)]+(k+1)[3^(k+1)]
=(3/4)[1+(2k-1)(3^k)+4(k+1)(3^k)]
=(3/4)[1+(3^k)(2k-1+4k+4)]
=(3/4)[1+(3^k)(6k+3)]
=(3/4){1+[3^(k+1)](2k+1)}

RHS=(3/4){1+[2(k+1)-1](3^k)}
=(3/4)[1+(2k+2-1)[3^(k+1)]
=(3/4)[1+(2k+1)[3^(k+1)]

LHS=RHS

Answer 3

n=k+1,

LHS
=1*3+2*(3^2)+...+k*(3^k)+(k+1)(3^(k+1))
= 3/4[1+(2k-1)(3^k)] + (k+1)(3^(k+1))
= 3/4[1+(2k-1)(3^k)] + (4/3)(k+1)(3^(k+1))]
= 3/4[1+(2k-1)(3^k)] + 4(k+1)(3^k)]
= 3/4[1+(6k+3)(3^k)]
= 3/4[1+(2k+1)(3^(k+1))]
= 3/4[1+(2(k+1)-1)) (3^(k+1))]
=RHS