By setting x equal to the appropriate values in the binomial expansion ( or one of its derivatives) evaluate the following : sum(k=1 to n) (n,k) k^2
To evaluate the sum \( \sum_{k=1}^{n} \binom{n}{k} k^2 \) using the binomial expansion, we can start by writing the binomial expansion formula:
\[ (a + b)^n = \binom{n}{0} a^{n} b^{0} + \binom{n}{1} a^{n-1} b^{1} + \binom{n}{2} a^{n-2} b^{2} + \ldots + \binom{n}{n} a^{0} b^{n} \]
Now, let's rearrange the formula and substitute \( a = k \) and \( b = 1 \):
\[ (k + 1)^n = \binom{n}{0} k^{n} 1^{0} + \binom{n}{1} k^{n-1} 1^{1} + \binom{n}{2} k^{n-2} 1^{2} + \ldots + \binom{n}{n} k^{0} 1^{n} \]
Now, to evaluate the sum \( \sum_{k=1}^{n} \binom{n}{k} k^2 \), we need to focus on the term \( \binom{n}{k} k^2 \). We can rewrite it as:
\[ \binom{n}{k} k^2 = \binom{n}{k} k^{2} 1^{n-k} \]
Comparing this with the expanded form above, we see that it corresponds to the \( \binom{n}{k} k^{n-k} 1^{k} \) term. Therefore, we can conclude that:
\[ \binom{n}{k} k^2 = \binom{n}{n-k} (n-k)^k \]
Substituting this back into the sum, we have:
\[ \sum_{k=1}^{n} \binom{n}{k} k^2 = \sum_{k=1}^{n} \binom{n}{n-k} (n-k)^k \]
Now, to evaluate this sum, we can use the symmetry property of binomial coefficients:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Therefore, the sum simplifies to:
\[ \sum_{k=1}^{n} \binom{n}{k} k^2 = \sum_{k=1}^{n} \binom{n}{k} (n-k)^k \]
To find the exact numerical value for this sum, you can substitute \( n \) with the appropriate value and compute the sum using a suitable mathematical software or by hand.