Show n^th terms of the progression logx logx2 logx3 logx4 is n(n+1÷2)log x

Answer 1

Answer to the question

Answer 2

did you mean
logx, log(x^2), log(x^3), log(x^4), ...
= logx, 2logx, 3logx, 4logx ....

Now it looks like an arithmetic progression
with a = logx and d = logx

term(n) = a + (n-1)d
= logx + (n-1)logx
= logx + nlogx - logx
= nlogx

looks like that is not what you meant

If you meant:
logx, (logx)^2, (logx)^3, ...
then it becomes a geometric progression with
a = logx and r = logx

term(n) = a r^(n-1)
= logx(logx)^(n-1)
= (logx)^n
= nlogx

Interesting that we get the same answer, but still not your answer.
Perhaps you want the sum of n terms ????

Tired of guessing what you mean,
retype with proper use of brackets to eliminate ambiguity.

Answer 3

logx, 2logx, 3logx, 4logx = logx(1+2+3+...+n) = logx * n(n+1)/2