Series with binomial coefficients
Task number: 3419
Add:
Variant
\( \sum\limits_{k=0}^n k\binom{n}{k} \)
Hint
Try to interpret the additional multiplicative factor \(k\) such that in a chosen \(k\)-tuple we distinguish one of its elements.
Solution
The expression represents the number of choices of \(k\)-element subsets from \(n\) elements, where we then choose a single element from the chosen subset.
We may attain the same result by first choosing a particular element and then extending it with a \((k-1)\)-element subset.
So we get: \( \sum\limits_{k=0}^n k\binom{n}{k} =n \sum_{k=0}^{n-1} \binom{n-1}{k-1} =n2^{n-1} \)
An alternative, formal calculation:
\( \sum\limits_{k=0}^n k\binom{n}{k} =\sum\limits_{k=1}^n k\binom{n}{k} = \sum\limits_{k=1}^n k\frac{n!}{k!(n-k)!} = n \sum\limits_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} = n \sum\limits_{k=0}^{n-1} \frac{(n-1)!}{k!(n-1-k)!} = n 2^{n-1} \)Another way to derive this is by differentiation . From the binomial formula: \( \sum\limits_{k=0}^n \binom{n}{k} x^k = (1+x)^n \) by differentiating by \(x\) we obtain: \( \sum\limits_{k=0}^n \binom{n}{k}k x^{k-1} = n(1+x)^{n-1} \).
Now we need only substitute \(x=1\).
Answer
The sum of the series is \(n 2^{n-1}\).
Variant
\( \sum\limits_{k=0}^n k^2\binom{n}{k} \)
Solution
The expression corresponds to the number of choices of \(k\)-element subsets from \(n\) elements, where we then separately select two distinguished elements from the chosen subset (we may choose the same element twice).
We may obtain the same result by first choosing the distinguished elements and then extending them to a \(k\)-element subset. We must distinguish the case where we select the same element twice.
So we obtain: \( \sum\limits_{k=0}^n k^2\binom{n}{k} =n \sum\limits_{k=0}^{n-1} \binom{n-1}{k-1}+ n(n-1)\sum\limits_{k=0}^{n-2} \binom{n-2}{k-2} =n2^{n-1}+n(n-1)2^{n-2}=n(n+1)2^{n-2} \)
Answer
The sum of the series is \(n(n+1)2^{n-2}\).
Variant
\( \sum\limits_{k=0}^{2n} (-1)^k\binom{4n}{2k} \)
Hint
Try substituting the imaginary unit \(i\) into the binomial identity.
Solution
\( \sum\limits_{k=0}^{2n} (-1)^k\binom{4n}{2k}= \sum\limits_{k=0}^{2n} i^{2k}\binom{4n}{2k}= Re\left(\sum\limits_{k=0}^{4n} i^k\binom{4n}{k}\right)= Re\left((1+i)^{4n}\right)= (-4)^n \)
Answer
The sum of the series is \((-4)^n\).
Variant
\( \sum\limits_{k=0}^{2n-1} (-1)^k\binom{4n}{2k+1} \)
Hint
Use the substitution \(-1=i^2\).
Then extend the series with complex terms such that the real part remains unchanged.
Then further transform the series so that you can apply the binomial theorem to it.
Solution
\( \sum\limits_{k=0}^{2n-1} (-1)^k\binom{4n}{2k+1}= \sum\limits_{k=0}^{2n-1} i^{2k}\binom{4n}{2k+1}= Re\left(\sum\limits_{l=0}^{4n-2} i^{l}\binom{4n}{l+1}\right)= \)
… we add the term \(i^{4n-1}\binom{4n}{4n}\) whose real component is zero …
\( =Re\left(\sum\limits_{l=0}^{4n-1} i^{l}\binom{4n}{l+1}\right)= Im\left(i\sum\limits_{l=0}^{4n-1} i^{l}\binom{4n}{l+1}\right)= Im\left(\sum\limits_{l=0}^{4n-1} i^{l+1}\binom{4n}{l+1}\right)= Im\left(\sum\limits_{j=1}^{4n} i^{j}\binom{4n}{j}\right)= \)
… we add the term \(i^{0}\binom{4n}{0}\) whose complex component is zero …
\( =Im\left(\sum\limits_{j=0}^{4n} i^{j}\binom{4n}{j}\right)= Im\left((1+i)^{4n}\right)= Im\left((-4)^n\right)=0 \)
Answer
The sum of the series is 0.
