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Answer: \( \displaystyle 100\)



---




Why?


The expression inside the set is




[
\frac(k+1)\,k\,k!k!
]



Because \(k!\) appears in both the numerator and the denominator it cancels:



[
\frac(k+1)\,k\,k!k!= (k+1)\,k
]



So for any integer \(k \ge 0\)



[
(k+1)\,k = k^2 + k
]



The set consists of all such products as \(k\) ranges over the non‑negative integers.

When you list a few values you see:




\(k\) \((k+1)\,k\)


0 0


1 2


2 6


3 12


4 20


These are exactly the numbers you wrote down.

There is no single "solution" number; it’s a sequence of values, each produced by plugging in a different \(k\).




If you had a specific value for \((k+1)\,k\) and wanted to find that particular \(k\), you would solve the quadratic equation:



[
k^2 + k - N = 0
]



with \(N\) being your target number. The positive root gives the required \(k\).



So in short: what you’re seeing are all the possible results of the expression \((k+1)\,k\), not a unique answer.
We need to craft final response: It's about interpreting "k+1 k" maybe product? Provide explanation that (k+1) k = k^2 + k. If we have equation (k+1)k = N; solve for k by quadratic formula.



But the user didn't specify the question exactly. We need to produce a comprehensive answer: show steps, examples, how to find k if given result value or expression.



Let's craft final answer:



Explain that k+1 times k equals k^2 + k. Provide algebraic simplification. Then talk about solving for k in terms of some number n; use quadratic formula. Provide demonstration with examples (e.g., 12 = k(k+1) => k=3). Mention negative solutions as well.



Also discuss if the user is asking to expand expression: k^2 + k or to find its factorization etc.



Let's produce final answer.The product \(k\,(k+1)\)




When you see the expression



[
k \times (k+1)
]



you can think of it as "take a number \(k\) and multiply it by the next integer".
Mathematically this is simply



[
k(k+1)=k^2+k .
]



So the product expands to a quadratic polynomial in \(k\).



---




1. Expanding (multiplying out)


If you want the result as a standard algebraic expression:



[
\boxedk(k+1)=k^2+k
]



That is, square the first factor (\(k^2\)) and add one copy of the first
factor (\(+k\)).



---




2. Evaluating for a particular \(k\)


If you know a specific value for \(k\), just plug it in:




\(k\) \(k(k+1)\)


0 \(0(1)=0\)


1 \(1(2)=2\)


2 \(2(3)=6\)


… …


---




3. Key Takeaway




The product \(k(k+1)\) is always a product of two consecutive integers*.


Algebraically, it simplifies to the quadratic expression \(\boxedk^2+k\).


Numerically, compute by multiplying \(k\) and \((k+1)\).



Feel free to experiment with different values of \(k\) or ask if you'd like to see more examples or explore related concepts!

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