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

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How do you expand ${(1 + 2 x)}^{6}$ $(1+2x)^6$ 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 $2$ $2$ to find:

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

Explanation:

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

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

$1, 6, 15, 20, 15, 6, 1$ $1, 6, 15, 20, 15, 6, 1$

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

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

to get:

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

Hence:

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

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

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

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