The sum basis binomial theorem in one variable for positive exponent can be expanded in the following two mathematical forms.
$(1).\,$ $(x+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^n y^0$ $+$ $\displaystyle \binom{n}{1} x^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} x^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^{0} y^n$
$(2).\,$ $(x+y)^n$ $\,=\,$ $^nC_0\,x^n y^0$ $+$ $^nC_1\,x^{n-1} y^1$ $+$ $^nC_2\,x^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,x^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,x^{0} y^n$
You can follow any one of the above expansions to prove the binomial theorem in one variable.
The Binomial theorem is defined in terms of two variables $x$ and $y$. In this case, the binomial theorem is expressed in one variable. So, substitute $x \,=\, 1$.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} (1)^n y^0$ $+$ $\displaystyle \binom{n}{1} (1)^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} (1)^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} (1)^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} (1)^{0} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\,(1)^n y^0$ $+$ $^nC_1\,(1)^{n-1} y^1$ $+$ $^nC_2\,(1)^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,(1)^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,(1)^{0} y^n$
Now, simplify the expansion of binomial theorem in one variable.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} \times 1 \times y^0$ $+$ $\displaystyle \binom{n}{1} \times 1 \times y^1$ $+$ $\displaystyle \binom{n}{2} \times 1 \times y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} \times 1 \times y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} \times 1 \times y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, \times 1 \times y^0$ $+$ $^nC_1\, \times 1 \times y^1$ $+$ $^nC_2\, \times 1 \times y^2$ $+$ $\cdots$ $+$ $^nC_r\, \times 1 \times y^r$ $+$ $\cdots$ $+$ $^nC_n\, \times 1 \times y^n$
Finally, the binomial theorem in one variable is written in the following forms.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} y^0$ $+$ $\displaystyle \binom{n}{1} y^1$ $+$ $\displaystyle \binom{n}{2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, y^0$ $+$ $^nC_1\, y^1$ $+$ $^nC_2\, y^2$ $+$ $\cdots$ $+$ $^nC_r\, y^r$ $+$ $\cdots$ $+$ $^nC_n\, y^n$
The expansion of the Binomial Theorem in one variable is derived in terms of $y$ but we are used to express it in terms of $x$. So, write the binomial theorem in one variable in terms of $x$ by replacing $y$ with $x$.
$(1).\,$ $(1+x)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^0$ $+$ $\displaystyle \binom{n}{1} x^1$ $+$ $\displaystyle \binom{n}{2} x^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^n$
$(2).\,$ $(1+x)^n$ $\,=\,$ $^nC_0\, x^0$ $+$ $^nC_1\, x^1$ $+$ $^nC_2\, x^2$ $+$ $\cdots$ $+$ $^nC_r\, x^r$ $+$ $\cdots$ $+$ $^nC_n\, x^n$
Find the values of binomial coefficients to substitute them in the expansion of the binomial theorem.
$\displaystyle \binom{n}{0}$ $\,=\,$ $^nC_0$ $\,=\,$ $\dfrac{n!}{0!(n-0)!}$ $\,=\,$ $1$
$\displaystyle \binom{n}{1}$ $\,=\,$ $^nC_1$ $\,=\,$ $\dfrac{n!}{1!(n-1)!}$ $\,=\,$ $\dfrac{n}{1!}$ $\,=\,$ $n$
$\displaystyle \binom{n}{2}$ $\,=\,$ $^nC_2$ $\,=\,$ $\dfrac{n!}{2!(n-2)!}$ $\,=\,$ $\dfrac{n(n-1)}{2!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{r}$ $\,=\,$ $^nC_r$ $\,=\,$ $\dfrac{n!}{r!(n-r)!}$ $\,=\,$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{n}$ $\,=\,$ $^nC_n$ $\,=\,$ $\dfrac{n!}{n!(n-n)!}$ $\,=\,$ $1$
Now, substitute the values of binomial coefficients in the expansion of the binomial theorem.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times x^0$ $+$ $n \times x^1$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
According to the zero power rule, the $x$ raised to the power zero is equal to one. The $x$ raised to the power one is denoted by $x$ in mathematics.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times 1$ $+$ $n \times x$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}\,x^r$ $+$ $\cdots$ $+$ $x^n$
Thus, the sum based binomial theorem in one variable is derived in mathematics and it is also written simply as follows.
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}\,x^3$ $+$ $\cdots$
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