Question

This problem has two parts, Part A and Part B.

Part A: Expand (x+1)6 using the Binomial Theorem.

Part B: Which method(s) could help identify the correct answer to Part A?

Select one answer for Part A and select all answers that apply for Part B.

B: ∑6k=0(6k)x6−k1k=(60)x6−010+(61)x6−111+(62)x6−212+(63)x6−313+(64)x6−414+(65)x6−515+(66)x6−616
A: x6+6x5+15x4+20x3+15x2+6x+1
B: ∑6k=1(6k)x6−k1k=(61)x6−111+(62)x6−212+(63)x6−313+(64)x6−414+(65)x6−515+(66)x6−616
B: The row corresponding to (a+b)6 in Pascal’s Triangle has the terms 1, 6, 15, 20, 15, 6, and 1.
A: x6+5x5+15x4+25x3+15x2+5x+1
B: The row corresponding to (a+b)6 in Pascal’s Triangle has the terms 1, 5, 15, 25, 15, 6, and 1.
A: x6−6x5+15x4+20x3+15x2−6x−1
A: x6+6x5+10x4+25x3+10x2+6x+1
B: ∑6k=0(6k)x6−k1k=(60)x016−0+(61)x116−1+(62)x216−2+(63)x316−3+(64)x416−4+(65)x516−5+(66)x616−6
B: The row corresponding to (a+b)6 in Pascal’s Triangle has the terms 1, 6, 10, 25, 10, 6, and 1.

Answer 1

Hard to read all those choices

in effect, (x+1)^6
= x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1

the row in Pascal's triangle I used was 1 6 15 20 15 6 1

I also see (a+b)^6 , which would be
= a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + 6a b^5 + b^6