The sum of $n$ terms of the series $\sum _{ x=1 }^{ n }{ \log { \cfrac { { 2 }^{ x } }{ { 3 }^{ x-1 } } } } $ is
- $\log { { \left( \cfrac { { 2 }^{ n-1 } }{ { 3 }^{ n+1 } } \right) }^{ n/2 } } $
- $\log { { \left( \cfrac { { 2 }^{ n+1 } }{ { 3 }^{ n-1 } } \right) }^{ n/2 } } $
- $\log { { \left( \cfrac { { 3 }^{ n-1 } }{ { 2 }^{ n+1 } } \right) }^{ n/2 } } $
- $\log { { \left( \cfrac { { 3 }^{ n+1 } }{ { 2 }^{ n-1 } } \right) }^{ n/2 } } $
Sum = log(2^1/3^0) + log(2^2/3^1) + ... + log(2^n/3^(n-1)) = log( (2^1 * 2^2 * ... * 2^n) / (3^0 * 3^1 * ... * 3^(n-1)) ). Numerator exponent = n(n+1)/2. Denominator exponent = (n-1)n/2. Sum = log( 2^(n(n+1)/2) / 3^(n(n-1)/2) ). This simplifies to log( (2^(n+1) / 3^(n-1))^(n/2) ).
Use the logarithm quotient rule to separate the general term into x times log 2 minus (x - 1) times log 3. Applying the summation formulas yields (n times (n + 1) divided by 2) times log 2 minus (n times (n - 1) divided by 2) times log 3. By combining these into a single logarithm with an exponent of n divided by 2, the sum equals the logarithm of 2 to the power of (n + 1) divided by 3 to the power of (n - 1), all raised to the power of n divided by 2. The result is the logarithm of the quantity 2 to the power of (n + 1) divided by 3 to the power of (n - 1), raised to the power of n divided by 2.