1+ 1/1+2+ 1/1+2+3....to n terms.
since 1+2+...+k = k(k+1)/2
Your sum is
n
∑ 2/(k(k+1))
k=1
see what you can do with that.
Note that
1/(k(k+1)) = 1/k - 1/k+1
sir answer this question 2n/n+1
what question? You have the sum correct.
To find the sum of the series 1 + 1/1 + 2 + 1/1 + 2 + 3 + ... up to n terms, we need to break down the series and apply the concept of partial sums.
First, let's write the series in a more organized form:
1 + (1/1 + 2) + (1/1 + 2 + 3) + ...
Now, let's focus on the pattern within parentheses. Each term within parentheses represents the sum of the first n natural numbers. We can find the sum of the first n natural numbers using the formula:
Sum = n * (n + 1) / 2
So, the sum within parentheses is:
(1 + 2 + 3 + ... + n) = n * (n + 1) / 2
Now, let's substitute this expression back into the original series:
1 + n/2 * (n + 1) / 2 + ...
We can see that the series consists of the term 1, followed by a term proportional to n^2. However, to determine the sum of the entire series, we need to find the sum of this pattern up to n terms.
The sum of a series with a pattern of terms can be calculated using the formula:
Sum = first term * (1 - common ratio ^ number of terms) / (1 - common ratio)
In this case, the first term is 1, and the common ratio is n/2 * (n + 1) / 2.
Using this formula, the sum of the entire series up to n terms is:
Sum = 1 * (1 - (n/2 * (n + 1) / 2) ^ n) / (1 - (n/2 * (n + 1) / 2))
Please note that simplifying this expression further would require specific values for n.