Show that the sum of nterm of the sequence logx,logx2, logx3 ,logx4.....given by n(n+1)/2. Logx
log x^n = n logs
so the sum is just
logx + 2logx + ... + nlogs
= (1+2+...+n)logx
look familiar yet?
the answer to this question.
To find the sum of the first n terms of the sequence logx, logx2, logx3, logx4..., we need to use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is:
Sn = (n/2) * (a + l)
Where:
- Sn is the sum of the n terms,
- n is the number of terms,
- a is the first term, and
- l is the last term.
In this case, we can see that the first term (a) is logx, and the last term (l) is logxn, where n is the nth term.
Since the terms of the given sequence are logx, logx2, logx3, logx4,..., we can see that the last term will be logxn, where n is the nth term.
So, the sum of the n terms is:
Sn = (n/2) * (logx + logxn)
Next, we can simplify the expression by using the logarithmic property that states:
loga + logb = log(ab)
Using this property, we can simplify the above expression:
Sn = (n/2) * logx + logxn
= (n/2) * logx + log(x^n)
Now, we can use another logarithmic property that states:
loga^n = n * loga
Using this property, we can further simplify the expression:
Sn = (n/2) * logx + n * logx
Factoring out logx, we have:
Sn = (n/2 + n) * logx
Finally, simplifying the expression further, we get:
Sn = (n(n+1)/2) * logx
Therefore, the sum of the first n terms of the sequence logx, logx2, logx3, logx4... is given by:
Sn = (n(n+1)/2) * logx
To show that the sum of the nth term of the sequence logx, logx2, logx3, logx4, ... is given by n(n+1)/2 * logx, we can use the principle of mathematical induction.
Step 1: Base Case
First, we need to verify that the formula holds for the first term, n = 1.
When n = 1, the sequence becomes logx.
And the sum of the first term, S(1), is logx.
Step 2: Inductive Hypothesis
Assume that the formula holds for some value k, where k is an arbitrary positive integer.
That means, the sum of the first k terms is given by S(k) = k(k+1)/2 * logx.
Step 3: Inductive Step
Now, we need to prove that the formula also holds for the next term, k+1.
For this, we will consider the sum of the first k+1 terms, which can be written as:
S(k+1) = logx + logx2 + logx3 + ... + logxk + logx(k+1)
Using the assumption from the inductive hypothesis:
S(k+1) = S(k) + logx(k+1)
S(k+1) = k(k+1)/2 * logx + logx(k+1)
Step 4: Proof
Now, let's simplify the expression for S(k+1):
S(k+1) = k(k+1)/2 * logx + logx(k+1)
S(k+1) = (k(k+1) + 2(k+1))/2 * logx
S(k+1) = (k^2 + 3k + 2)/2 * logx
S(k+1) = (k+1)(k+2)/2 * logx
S(k+1) = (k+1)((k+1)+1)/2 * logx
This shows that the formula holds for the (k+1)th term as well.
Step 5: Conclusion
By using mathematical induction, we have shown that the sum of the nth term of the sequence logx, logx2, logx3, logx4, ... is given by n(n+1)/2 * logx.
Therefore, the sum of the first n terms of the sequence can be expressed as S(n) = n(n+1)/2 * logx.