How do you expand #(1+2x)^6# using Pascal’s Triangle?

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Answer 1 · 2016-01-17T22:35:03

Use the appropriate row of Pascal's triangle and a sequence of powers of #2# to find:

#(1+2x)^6 = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6#

Write out Pascal's triangle as far as the row which begins #1, 6#...

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This gives you the sequence of coefficients for #(a+b)^6#:

#1, 6, 15, 20, 15, 6, 1#

Then we can account for the factor of #2# of the #2x# term, by multiplying by a sequence of powers of #2#:

#1, 2, 4, 8, 16, 32, 64#

to get:

#1, 12, 60, 160, 240, 192, 64#

Hence:

#(1+2x)^6 = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6#