Question

Prove using mathematical induction that \(\sum\limits_{j = 2}^n C (j,2) = C(n + 1,3)\) whenever \(n\) is an integer greater than\(1\).

Short answer

Use a proof by induction.

Step-by-step solution

definition mathematical induction

For each and every natural number n, mathematical induction is a strategy for proving a statement, theorem, or formula that is thought to be true.

Solution

Let us simplify,

For\(n = 2\), sum_ {\(\sum\limits_{j = 2}^n {\left( {\begin{array}{*{20}{l}}j\\2\end{array}} \right)} = 1 = \left( {\begin{array}{*{20}{c}}{2 + 1}\\3\end{array}} \right)\)

Suppose it is true for\(n - 1\), i.e.\(\sum\limits_{j = 2}^{n - 1} {\left( {\begin{array}{*{20}{l}}j\\2\end{array}} \right)} = \left( {\begin{array}{*{20}{l}}n\\3\end{array}} \right)\).

Then\(\sum\limits_{j = 2}^n {\left( {\begin{array}{*{20}{l}}j\\2\end{array}} \right)} = \sum\limits_{j = 2}^{n - 1} {\left( {\begin{array}{*{20}{l}}j\\2\end{array}} \right)} + \left( {\begin{array}{*{20}{l}}n\\2\end{array}} \right) = \left( {\begin{array}{*{20}{l}}n\\3\end{array}} \right) + \left( {\begin{array}{*{20}{l}}n\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{n + 1}\\3\end{array}} \right).\)

Therefore, Use an inductive proof.