How do you expand $(x + 2)^5$ using Pascal’s Triangle?

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Answer 1 · 2015-12-31T07:07:18

$x^5+10x^4+40x^3+80x^2+80x+32$

The $5"th"$ row of Pascal's triangle is

$1,5,10,10,5,1$

These values are the coefficients in a binomial expansion to the $5"th"$ power.

$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$

Notice the pattern of the exponents: the exponent of $a$ starts at $5$ and goes to $0$ , and $b$ starts at $0$ and increases to $5$ .

Apply the rule to $(x+2)$ :

$(x+2)^5=x^5+5x^4(2)+10x^3(2^2)+10x^2(2^3)+5x(2^4)+2^5$

$=>x^5+10x^4+40x^3+80x^2+80x+32$

How do you expand (x + 2)^5(x + 2)^5 using Pascal’s Triangle?