Show n^th terms of the progression logx logx2 logx3 logx4 is n(n+1÷2)log x
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 to the question
logx, 2logx, 3logx, 4logx = logx(1+2+3+...+n) = logx * n(n+1)/2
To find the nth term of the given progression, we need to first understand the pattern in the sequence. Let's analyze the given sequence: logx, logx2, logx3, logx4.
We can observe that each term in the sequence is the logarithm of x with an exponent that corresponds to the position of the term in the sequence.
The first term, logx, corresponds to n = 1.
The second term, logx2, corresponds to n = 2.
The third term, logx3, corresponds to n = 3.
The fourth term, logx4, corresponds to n = 4.
So, we can conclude that for the nth term, the exponent is equal to n.
Now, to find the formula for the nth term, let x be the base of the logarithm, and let y be the nth term. We have:
y = logx^n
Using the rule of logarithms, we can write this as:
y = n * logx
Therefore, the formula for the nth term in the given progression is:
n * logx
In your question, you also mention (n+1÷2). It seems there is some ambiguity in the expression. If you clarify what you mean by that, I can help you further.