Question on combinations . Application of it on Pascal's Triangle to prove an inductive property.

Question

Prove that each number A in pascal's triangle is equal to the sum of the numbers in the previous right diagonal starting from its leftmost number through the number which is located in the same left diagonal as A.

Answer 1

Hint:

The bottom dot consists of 2 dots from diagonally above, which is the rule of how Pascal's triangle is formed. Try finding what the top left dot is made of according to the question. Notice anything?

Spoiler:

Let a endless diagonal, shown in the question above, be called $n_1,n_2,n_3,...$ from the top left to going down infinitely. Let the diagonal one down from it be called $m_1,m_2,m_3,...$ The question states that $m_p = \sum_{i=1}^{p} n^p$.

Lemma1: $m_i=m_{i-1}+n_i$. This is obviously true as that is how Pascal's triangle is built.

Lemma2: $m_1$ and $n_1$ = 1. This is, once again, trivial.

Answer (built on from spoiler):

We will now prove it by induction. The base case, $m_2$, should be $n_1+n_2 = m_1+n_2$, which fits Lemma1. Therefore, the base case is true.

Assume that $m_k$ works, and is equal to $\sum_{i=1}^{k} n^i$. Then, $m_{k+1} = m_k + n_{k+1}$. This gives us $m_{k+1} = \sum_{i=1}^{k} n^i + n^{k+1} = \sum_{i=1}^{k+1} n^i$, which completes the proof.