Question

question 5 (1 point) saved
express as a single term from pascals triangle: $t_{n + 2,r+3}-t_{n + 1,r+2}$
$t_{n+1,r + 3}$
$t_{n+2,r+2}$
$t_{n+3,r+3}$
$t_{n+1,r+2}$
$t_{n,r}$

Explanation:

Step1: Recall Pascal's identity

The Pascal's identity states that \(t_{n,k}=t_{n - 1,k}+t_{n - 1,k - 1}\), which can be rewritten as \(t_{n,k}-t_{n - 1,k - 1}=t_{n - 1,k}\).
Let \(n'=n + 2\) and \(k'=r + 3\), then \(t_{n'+2,k'}-t_{n'+1,k'-1}=t_{n'+1,k'}\). Substituting back \(n'=n + 2\) and \(k'=r + 3\), we have \(t_{n + 2,r+3}-t_{n + 1,r + 2}=t_{n + 1,r+3}\).

Answer:

\(t_{n + 1,r+3}\)