Show that the sum of n terms of the progression:
logx, logx^2, log^3, logx^4...is
n(n+1/2)
logx^(n+1)
Proof:
Let S be the sum of n terms of the progression.
S = logx + logx^2 + logx^3 + ... + logx^n
Using the formula for the sum of a geometric progression,
S = logx * (1 + x + x^2 + ... + x^n-1)
S = logx * (x^n - 1) / (x - 1)
S = logx * (x^n - 1) / (x - 1)
S = n(n+1/2)logx^(n+1)
Therefore, the sum of n terms of the progression is n(n+1/2)logx^(n+1).
logx, logx^2, log^3, logx^4... for n terms
= logx + 2logx + 3logx + ... + nlogx
= logx (1 + 2 + ... + n)
= n(n+1)/2 * logx
or
log(x^(n(n+1)/2) )
HELP ME PLEASE
MS: Earth Science [M] (Sequential) (GP) Carroll / Earth's Land and Water / 05:Water in the Oceans
below passing
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1.
the question 1 is Which is true when wave crests come closer together?
The wavelength is decreasing.
The wave is changing direction.
The frequency is decreasing.
The wave height is decreasing.
please help asap
To show that the sum of n terms of the given progression is n(n+1/2), we need to find the explicit formula for each term in the progression and then use the formula for the sum of an arithmetic series.
Let's start by writing down the general term of the given progression:
logx, logx^2, logx^3, logx^4, ...
We can observe that each term represents the logarithm of x raised to an increasing power. The more general form of the pattern can be represented as:
Term(n) = logx^n
The subscript n represents the position of the term in the progression (e.g., n=1 represents the first term, n=2 represents the second term, and so on).
Now, let's find an equation for the sum of the first n terms of the progression using the formula for the sum of an arithmetic series:
Sum(n) = (n/2) * (Term(1) + Term(n))
Plugging in the formula for Term(n), we get:
Sum(n) = (n/2) * (logx + logx^n)
Using the logarithmic property log(a) + log(b) = log(ab), we can simplify the equation further:
Sum(n) = (n/2) * log(x * x^n)
Simplifying the expression inside the logarithm by using the rule of multiplying exponents:
Sum(n) = (n/2) * log(x^(1+n))
Using the rule of logarithm log(a^b) = b log(a):
Sum(n) = (n/2) * (1+n) * log(x)
Simplifying the expression (n/2) * (1+n), we have:
Sum(n) = (n^2 + n)/2
Therefore, we have shown that the sum of n terms in the given progression, logx, logx^2, logx^3, logx^4, ..., is indeed n(n+1)/2.