a.maths1

Let P(x) be the proposition of
" 1 + 2x3 + 3x3^2 + 4x3^3 + ... + n(3^n-1) = 1/4 [ 1 + (2n - 1)3^n ] "
for all positive integers n.
when n = 1,LHS = 1(3^1 - 1) = 1(3 - 1) = 1(2) = 2
RHS = 1/4 [ 1+ ( 2(1) - 1 )3^1 ] = 1/4 (1 + 3) = 1/4 (4) = 1

LHS not equal to RHS , it can not prove - -"...

你是不是出錯題目呢?
左式同右式當n=1時,都不是相等的,
所以這數是計不到的

Let P(x) is the proposition that 1 + 2x3 + 3x3^2 + 4x3^3 + ... + n(3^n-1) = 1/4 [ 1 + (2n - 1)3^n ] for all positive integers n.
For n = 1,
L.H.S = 1
R.H.S = 1
Therefore, it is true for n = 1.
Assume that P(k) is true for some integer k>=1,
when n = k+1
1 + 2x3 + 3x3^2 + 4x3^3 + ... + k(3^k-1) + (k+1)(3^k)
= 1/4 [ 1 + (2k - 1)3^k ] + (k+1)3^k
= 1/4+(3^k)/4*[2k-1+4k+4]
= 1/4+(3^k)/4*3(2(k+1)-1)
=1/4[1+(2(k+1)-1)3^(k+1)]
Therefore, it is true for n = k+1.
By the principle of Mathematical Induction.
1 + 2x3 + 3x3^2 + 4x3^3 + ... + n(3^n-1) = 1/4 [ 1 + (2n - 1)3^n ] for all positive integers n.