%5E6-binomial-expansion.jpeg "(2+x)^6 Binomial Expansion")
Binomial Expansion of (2+x)^6
In algebra, the binomial expansion is a method of expanding an expression of the form $(a+b)^n$, where $a$ and $b$ are constants and $n$ is a positive integer. In this article, we will focus on the binomial expansion of $(2+x)^6$.
What is Binomial Expansion?
The binomial expansion is a mathematical formula that allows us to expand an expression of the form $(a+b)^n$ into a sum of terms involving various powers of $a$ and $b$. The formula is given by:
$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$
where $\binom{n}{k}$ is the binomial coefficient, which can be calculated using the formula:
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Binomial Expansion of (2+x)^6
Using the binomial expansion formula, we can expand $(2+x)^6$ as follows:
$(2+x)^6 = \sum_{k=0}^6 \binom{6}{k} 2^{6-k} x^k$
To evaluate this expression, we need to calculate the binomial coefficients and the powers of 2 and $x$ for each term.
Term 1: k = 0
$\binom{6}{0} = \frac{6!}{0!(6-0)!} = 1$ $2^{6-0} = 2^6 = 64$ $x^0 = 1$ So, the first term is $64$.
Term 2: k = 1
$\binom{6}{1} = \frac{6!}{1!(6-1)!} = 6$ $2^{6-1} = 2^5 = 32$ $x^1 = x$ So, the second term is $192x$.
Term 3: k = 2
$\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15$ $2^{6-2} = 2^4 = 16$ $x^2 = x^2$ So, the third term is $240x^2$.
Term 4: k = 3
$\binom{6}{3} = \frac{6!}{3!(6-3)!} = 20$ $2^{6-3} = 2^3 = 8$ $x^3 = x^3$ So, the fourth term is $160x^3$.
Term 5: k = 4
$\binom{6}{4} = \frac{6!}{4!(6-4)!} = 15$ $2^{6-4} = 2^2 = 4$ $x^4 = x^4$ So, the fifth term is $60x^4$.
Term 6: k = 5
$\binom{6}{5} = \frac{6!}{5!(6-5)!} = 6$ $2^{6-5} = 2^1 = 2$ $x^5 = x^5$ So, the sixth term is $12x^5$.
Term 7: k = 6
$\binom{6}{6} = \frac{6!}{6!(6-6)!} = 1$ $2^{6-6} = 2^0 = 1$ $x^6 = x^6$ So, the seventh term is $x^6$.
Final Result
Combining all the terms, we get:
$(2+x)^6 = 64 + 192x + 240x^2 + 160x^3 + 60x^4 + 12x^5 + x^6$
This is the binomial expansion of $(2+x)^6$.
