Chapter 12: Q. 33 (page 839)

Find the coefficient of $x^{2}$ in the expansion of ${(2 x + 1)}^{7}$ .

Short Answer

Expert verified

The coefficient of $x^{2}$ is $84 .$

Step by step solution

Given information:

Consider the expression ${(2 x + 1)}^{7}$ .

To find the coefficient of $x^{2}$

Formulas :

Formula 1: To find a particular term in an expression:

${(x + a)}^{n} = (\begin{matrix} n \\ n - j \end{matrix}) a^{(n - j)} x^{j}$

Formula 2: Binomial Coefficient:

$(\begin{matrix} n \\ j \end{matrix}) = \frac{n!}{j! \times (n - j)!}$

Step 3:

On comparing ${(2 x + 1)}^{7}$ with ${(x + a)}^{n}$ , we get,

$x = 2 x, a = 1, n = 7$

To find coefficient of $x^{2}$ , $\Rightarrow j = 2$ .

Substituting the values in the formula 1:

${(2 x + 1)}^{7} = (\begin{matrix} 7 \\ 7 - 2 \end{matrix}) {(1)}^{(7 - 2)} {(2 x)}^{2}$

Simplification:

${(2 x + 1)}^{7} = (\begin{matrix} 7 \\ 5 \end{matrix}) 1^{5} (2^{2} x^{2})$

$= 21 \times 4 x^{2}$

role="math" localid="1647343658235" $[U s e f o r m u l a 2 : (\begin{matrix} 7 \\ 5 \end{matrix}) = \frac{7!}{5! \times (7 - 5)!} = \frac{7!}{5! \times 2!} = \frac{7 \times 6 \times 5!}{5! \times (2 \times 1)} = \frac{7 \times 6}{2} = 21]$

$= 84 x^{2}$

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Find the coefficient of $x^{2}$ in the expansion of ${(2 x + 1)}^{7}$ .

Short Answer

Step by step solution

Given information:

Formulas :

Step 3:

Substituting the values in the formula 1:

Simplification:

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Find the coefficient of $x^{2}$ in the expansion of ${(2 x + 1)}^{7}$ .