show that the sum of nterm of the progression is logx,logx^2,logx^3,logx^4---is n(n+1)/2 logx
log x^n = n log x
a 2a 3a 4a .......
idek
Not well stated...
Want full workings...
what is the answer to the question
To prove that the sum of the nth term of the given progression is equal to (n(n+1)/2)logx, we can use the concept of arithmetic progression and the formula for the sum of the first n terms of an arithmetic progression.
The given progression is: logx, logx^2, logx^3, logx^4, ...
Notice that each term in the sequence is obtained by taking the logarithm of the term of x raised to the power of its position in the sequence.
Let's denote the nth term of the given progression as Tn. So, Tn = log(x^n).
Now, let's find the sum of the first n terms of the progression using the arithmetic progression sum formula:
Sn (sum of the first n terms) = (n/2)(2a + (n-1)d),
where a is the first term of the arithmetic progression and d is the common difference.
In our case, a = logx (the first term) and d = logx^2 - logx = log(x^2/x) = logx.
Substituting these values into the formula, we get:
Sn = (n/2)(2logx + (n-1)logx).
Simplifying further:
Sn = (n/2)(2 + n-1)logx,
Sn = (n/2)(n+1)logx.
Therefore, the sum of the nth term of the given progression is indeed (n(n+1)/2)logx, which completes the proof.