Show that the sum n terms of the progression:
logx, logx^2, logx^3,logx^4,... is
n[n+1/2]logx
To find the sum of the first n terms of the progression log(x), log(x^2), log(x^3), log(x^4), ... , we can rewrite each term as a multiple of log(x) and then apply the sum of an arithmetic series formula.
log(x) is the 1st term in the progression.
log(x^2) = 2 * log(x), which is the 2nd term.
log(x^3)= 3 * log(x), is the 3rd term.
log(x^4) = 4 * log(x), is the 4th term.
Thus, the n-th term will be:
n * log(x)
Let S_n be the sum of the first n terms, then S_n can be written as:
S_n = log(x) + 2log(x) + 3log(x) + ... + nlog(x)
S_n = log(x)(1 + 2 + 3 + ... + n)
S_n = log(x) * (n(n+1)/2)
Therefore, the sum of the first n terms of the progression log(x), log(x^2), log(x^3), log(x^4), ... is n[n+1/2]log(x)