Chapter 12: Q 20. (page 836)

Expand ${(x + 3)}^{5}$ using Binomial Theorem

Short Answer

Expert verified

The value of ${(x + 3)}^{5} = x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 243$

Step by step solution

Step 1. Defination

Let $x$ and $a$ be real numbers for any positive integer $n$ , we have ${(x + a)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a x^{n - 1} + \dots \dots + (\begin{matrix} n \\ j \end{matrix}) a^{j} x^{n - j} + \dots \dots + (\begin{matrix} n \\ n \end{matrix}) a^{n} = \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) x^{n - j} a^{j}$

Step 2. Explanation

$\Rightarrow {(x + 3)}^{5} = (\begin{matrix} 5 \\ 0 \end{matrix}) {(3)}^{0} {(x)}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(3)}^{1} {(x)}^{4} + (\begin{matrix} 5 \\ 2 \end{matrix}) {(3)}^{2} {(x)}^{3} + (\begin{matrix} 5 \\ 3 \end{matrix}) {(3)}^{3} {(x)}^{2} + (\begin{matrix} 5 \\ 4 \end{matrix}) {(3)}^{4} {(x)}^{1} + (\begin{matrix} 5 \\ 5 \end{matrix}) {(3)}^{5} {(x)}^{0}$

localid="1647227300943" $\Rightarrow {(x + 3)}^{5} = \frac{5!}{0! (5 - 0)!} x^{5} + \frac{5!}{1! (5 - 1)!} (3) x^{4} + \frac{5!}{2! (5 - 2)!} (9) x^{3} + \frac{5!}{3! (5 - 3)!} (27) x^{2} + \frac{5!}{4! (5 - 4)!} (81) x^{1} + \frac{5!}{5! (5 - 5)!} (243)$

$\Rightarrow {(x + 3)}^{5} = x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 243$

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Expand ${(x + 3)}^{5}$ using Binomial Theorem

Short Answer

Step by step solution

Step 1. Defination

Step 2. Explanation

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Expand ${(x + 3)}^{5}$ using Binomial Theorem