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Write the first four terms in the expansion of: (x+3)^17

Write the first four terms in the expansion of: (x+3)^17-example-1

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Using the binomial theorem:


(a+b)^n=\Sigma((n)/(i))a^(n-i)*b^i

In this case n=17; a=x and b=3, the sum starts in i=0 until n.

And :


\begin{bmatrix}{n} & {\placeholder{⬚}} \\ {i} & {\placeholder{⬚}}\end{bmatrix}=(n!)/((n-i)!*i!)

Substituing the values:

First term: i=0.


(x+3)^(17)=x^(17-0)*3^0*(17!)/((17-0)!*0!)=x^(17)*(17!)/(17!)=x^(17)

Remember that 0!=1.

Second term: i=1


x^(17-1)*3^1*(17!)/((17-1)!*1!)=x^(16)*3*(17*16!)/(16!)=x^(16)*3*17=51x^(16)

Third term: i=2.


\begin{gathered} x^(17-2)*3^2*(17!)/((17-2)!*2!)=x^(15)*9*(17*16*15!)/(15!*2!) \\ =x^(15)*9*(17*16)/(2)=1224x^(15) \end{gathered}

Answer:

x^17 + 51*x^16 + 1224*x^15.

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