Evaluate Finite Summation Expression

[bincy]

How to evaluate the following expression?\(\displaystyle \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} \)
regards,
Bincy

 

[Dustinsfl]

How to evaluate the following expression?\(\displaystyle \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} \)
regards,
Bincy

Have you tried writing out the first few terms and the last 2 terms?

 

[bincy]

But I didn't get the ans.

 

[chisigma]

How to evaluate the following expression?\(\displaystyle \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} \)

The explicit expression [if it exists...] of the finite sum may be [probably...] found using the so called 'Noerdlund- Rice Integral'...

$\displaystyle \sum_{j=\alpha}^{n} \binom{n}{j}\ (-1)^{j} f(j)= (-1)^{n} \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{z\ (z-1)\ (z-2)...(z-n)}\ dz$ (1)

... setting $\alpha=0$, $\displaystyle f(z)=\frac{1}{(2+z)^{k}}$ and with proper choice of the path $\gamma$. The details are quite complex and require more work...

Kind regards

$\chi$ $\sigma$

 

[blue_raver22]

I would approach evaluating this finite summation expression by first understanding its components. The expression contains a summation symbol, denoted by the capital Greek letter sigma, which indicates that we need to add up a series of terms. The index i represents the variable that we will be summing over, starting at i=0 and ending at i=N. The expression also contains a binomial coefficient, denoted by the parentheses with N and i inside, which represents the number of ways to choose i objects from a set of N objects. Additionally, the expression contains a power of -1, which alternates between positive and negative values, and a fraction with a variable k in the exponent.

To evaluate this expression, we can use the concept of mathematical induction. We start by plugging in i=0 into the expression and then gradually increase i until we reach i=N. By doing so, we will have evaluated the expression for all values of i from 0 to N, and then we can add up all these values to get the final result.

In the first step, when i=0, the expression becomes \binom{N}{0} \left(-1\right)^{0}\left(\frac{1}{2+0}\right)^{k} = 1\cdot 1 \cdot \left(\frac{1}{2}\right)^{k} = \left(\frac{1}{2}\right)^{k}.

In the second step, when i=1, the expression becomes \binom{N}{1} \left(-1\right)^{1}\left(\frac{1}{2+1}\right)^{k} = N\cdot (-1) \cdot \left(\frac{1}{3}\right)^{k} = -N\cdot \left(\frac{1}{3}\right)^{k}.

Similarly, we can continue this process until we reach the final step when i=N. At this point, the expression becomes \binom{N}{N} \left(-1\right)^{N}\left(\frac{1}{2+N}\right)^{k} = 1\cdot (-1)^{N} \cdot \left(\frac{1}{N+2}\right)^{k} = (-1)^{N} \cdot \left(\frac{1}{N+2}\right)^