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# Prove that displaystylesum_{i=1}^{n} left(begin{array}{c}i 2end{array}right)=left(begin{array}{c}n+1 3end{array}right) forall ngeq2 # Prove that displaystylesum_{i=1}^{n} left(begin{array}{c}i 2end{array}right)=left(begin{array}{c}n+1 3end{array}right) forall ngeq2

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Probability and combinatorics asked 2020-11-07
Prove that $$\displaystyle\sum_{i=1}^{n} \left(\begin{array}{c}i\\ 2\end{array}\right)=\left(\begin{array}{c}n+1\\ 3\end{array}\right)$$
$$\forall n\geq2$$

## Answers (1) 2020-11-08

Step 1. Definitions
Definition combination
$$nCr=\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$$
with $$n!=n\times(n-1)\times\dotsb\times2\times1$$.
Pascal's equation
$$\left(\begin{array}{c}n+1\\ k\end{array}\right)=\left(\begin{array}{c}n\\ k\end{array}\right)+\left(\begin{array}{c}n\\ k-1\end{array}\right)$$
Step 2. Solution
To proof: $$\displaystyle\sum_{i=1}^{n} \left(\begin{array}{c}i\\ 2\end{array}\right)=(n+1)$$ for all positive integers.
Proof by induction
Let P(n) be the statement "$$\displaystyle\sum_{i=1}^{n} \left(\begin{array}{c}i\\ 2\end{array}\right)=(n+1)$$".
Basis step. $$n=2$$
$$\displaystyle\sum_{i=1}^{n} \left(\begin{array}{c}i\\ 2\end{array}\right)=\displaystyle\sum_{i=1}^{2} \left(\begin{array}{c}i\\ 2\end{array}\right)=\left(\begin{array}{c}i\\ 2\end{array}\right)+\left(\begin{array}{c}2\\ 2\end{array}\right)=0+1=1$$
$$\left(\begin{array}{c}n+1\\ 3\end{array}\right)=\left(\begin{array}{c}2+1\\ 3\end{array}\right)=\left(\begin{array}{c}3\\ 3\end{array}\right)=1$$
Thus P(2) is true.
Inductive step. Let P(k) be true.
$$\displaystyle\sum_{i=1}^{k} \left(\begin{array}{c}i\\ 2\end{array}\right)=\left(\begin{array}{c}k+1\\ 3\end{array}\right)$$
We need to proof that $$P(k+1)$$ is true.
$$\displaystyle\sum_{i=1}^{k+1} \left(\begin{array}{c}i\\ 2\end{array}\right)$$
$$=\displaystyle\sum_{i=1}^{k+1} \left(\begin{array}{c}i\\ 2\end{array}\right)+\left(\begin{array}{c}k+1\\ 2\end{array}\right)$$
$$\left(\begin{array}{c}k+1\\ 3\end{array}\right)+\left(\begin{array}{c}k+1\\ 2\end{array}\right) \ \text{Since P(k) is true}$$
$$=\left(\begin{array}{c}(k+1)+1\\ 3\end{array}\right) \ \text{Pascal's identity}$$
Thus $$P(k+1)$$ is true.
Conclution. By the principle of methematical induction, P(n) is true for all positive integers n.

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