\(1 = 1\)

\(1 + 1 = 2\)

\(1 + 2 + 1 = 4\)

\(1 + 3 + 3 + 1 = 8\)

\(1 + 4 + 6 + 4 + 1 = 16\)

a) Without doing the computation, explain why the sum of entries in any row shown above will turn out to be double the sum of the entries in the row just above it.

HINT: Each entry in any one row is the sum of two entries in the row above it.

b) Explain why the sum of entries in the \(n\)th row is sure to be \(2^{n}\). (Remember to call the top row of Pascal’s triangle the zero-th row.)

\(1 – 1 = 0\)

\(1 – 2 + 1 = 0\)

\(1 – 3 + 3 – 1 = 0\)

\(1 – 4 + 6 – 4 + 1 = 0\)

\(1 – 5 + 10 – 10 + 5 – 1 = 0\)