Question

How do you use the binomial series to expand ${(x + 2)^7}$ ?

Answer 1

First, we need to know about the relation between binomial expansion and Pascal’s triangle. The word binomial stands for expressions having two terms. For example: $(1 + x)$ , $(x + y)$ and etc.
Pascal triangle is an arrangement of the numbers ${}^n{C_r}$ in a triangular form.
Binomial expansion of ${(a + b)^n}$ is
${(a + b)^n} = {}^n{C_0}{a^n}{b^o} + {}^n{C_1}{a^{n - 1}}{b^1} + \cdots + {}^n{C_r}{a^{n - r}}{b^r} + \cdots + {}^n{C_n}{a^0}{b^n}$ .
For example, from the fifth row we can write down the expansion of ${(a + b)^4}$ and from the sixth row we can write down the expansion of ${(a + b)^5}$ and so on. We know the terms (without coefficients) of ${(a + b)^5}$ are:
${a^5},{a^4}b,{a^3}{b^2},{a^2}{b^3},a{b^4},{b^5}$
The two variable’s power adding them together
And the sixth row of the Pascal triangle is
$1$ $5$ $10$ $10$ $5$ $1$
Using these two we can write
${(a + b)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5}$

Complete step-by-step solution:
The given binomial expansion is ${(x + 2)^7}$ .
Let consider, $a = x$ and $b = 2$
$n = 7$
The seventh row of Pascal triangle is
$1$ $7$ $21$ $35$ $35$ $21$ $7$ $1$
The terms (without coefficients) of ${(a + b)^7}$
${a^7},{a^6}b,{a^5}{b^2},{a^4}{b^3},{a^3}{b^4},{a^2}{b^5},a{b^6},{b^7}$
The two variable’s power adding them together
Using the two, we can write,
${(a + b)^7} = {a^7} + 7{a^6}b + 21{a^5}{b^2} + 35{a^4}{b^3} + 35{a^3}{b^4} + 21{a^2}{b^5} + 7a{b^6} + {b^7}$
Then $a$ and $b$ value substituting $x$ and $7$ respectively in the equation of ${(a + b)^7}$
This can be applied
${(x + 2)^7} = {x^7} + 7{x^6}(2) + 21{x^5}{(2)^2} + 35{x^4}{(2)^3} + 35{x^3}{(2)^4} + 21{x^2}{(2)^5} + 7x{(2)^6} + {(2)^7}$
Now, we have to find the value of every required power of $(2)$ . For example, ${(2)^2} = 2 \times 2 = 4$
And finding every power of values,
${(x + 2)^7} = {x^7} + 7{x^6}(2) + 21{x^5}(4) + 35{x^4}(8) + 35{x^3}(16) + 21{x^2}(32) + 7x(64) + (128)$
Now we will multiply the coefficient of $x$ by the power of $(2)$ values.
$7{x^6}(2) = 7 \times 2{x^6} = 14{x^6}$
$21{x^5}{(2)^2} = 21 \times 4{x^5} = 84{x^5}$
$35{x^4}{(2)^3} = 35 \times 8{x^4} = 280{x^4}$
$35{x^3}{(2)^4} = 35 \times 16{x^3} = 560{x^3}$
$21{x^2}{(2)^5} = 21 \times 32{x^2} = 672{x^2}$
$7x{(2)^6} = 7 \times 64x = 448x$
Now, substituting in the equation, we get,
${(x + 2)^7} = {x^7} + 14{x^6} + 84{x^5} + 280{x^4} + 560{x^3} + 672{x^2} + 448x + 128$

Then, the required ${(x + 2)^7}$ binomial expansion is
${(x + 2)^7} = {x^7} + 14{x^6} + 84{x^5} + 280{x^4} + 560{x^3} + 672{x^2} + 448x + 128$

Note: If $n$ is any positive integer, then
${(a + b)^n} = {}^n{C_0}{a^n}{b^o} + {}^n{C_1}{a^{n - 1}}{b^1} + \cdots + {}^n{C_r}{a^{n - r}}{b^r} + \cdots + {}^n{C_n}{a^0}{b^n}$, $n \in \mathbb{N}$
Some cases of binomial theorem,
Replacing $b$ by $( - b)$, in the binomial expansion of ${(a + b)^n},n \in \mathbb{N}$ , we get
${(a - b)^n} = {}^n{C_0}{a^n}{b^o} - {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} - \cdots + {( - 1)^r}{}^n{C_r}{a^{n - r}}{b^r} + \cdots + {( - 1)^n}{}^n{C_n}{a^0}{b^n}$
Observe that the sign $' + '$ and $' - '$ appear alternately in the binomial expansion of ${(a - b)^n}$ .
Replacing $a$ by $1$ and $b$ by $x$ , in the binomial expansion of ${(a + b)^n}$ , we get
${(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + \cdots + {}^n{C_r}{x^r} + \cdots + {}^n{C_n}{x^n}.$