無窮級數的和的一個數學問題

求 sigma(n=1 to無限大)_(1+2+2^2+2^3+…+2^n)/3^n
Sol
1+2+2^2+2^3+…+2^n
=1+1+2+2^2+2^3+…+2^n－1
=2^(n+1)－1
sigma(n=1 to 無限大)_(1+2+2^2+2^3+…+2^n)/3^n
=sigma(n=1 to 無限大)_(2^(n+1)－1)/3^n
=sigma(n=1 to 無限大)_2^(n+1)/3^n－sigma(n=1 to 無限大)_1/3^n
=2sigma(n=1 to 無限大)_2^n/3^n－sigma(n=1 to 無限大)_(1/3)^n
=2 sigma(n=1 to 無限大)_(2/3)^n－sigma(n=1 to 無限大)_(1/3)^n
=2*[2/3/(1－2/3)]-[(1/3)/(1－1/3)]
=2*[2/(3－2)]－[1/(3－1)]
=4－1/2
=7/2

1+2+...+2^n

=(2^(n+1)-1)/(2-1)

=2^(n+1)-1

Σ (1+2+...+2^n)/3^n

=Σ (2^(n+1)-1)/3^n

=2Σ (2/3)^n - Σ (1/3)^n

= 4-1/2

=7/2

1+2+...+2^n = 2^(n+1) - 1

Σ (1+2+...+2^n)/3^n
= Σ [2^(n+1)-1]/3^n
= Σ [2*2^n-1)/3^n
= 2*Σ (2/3)^n - Σ (1/3)^n
= 4 - 1/2
= 7/2