Math Doubts

Proof of Sum basis Binomial Theorem in One variable for Positive exponent

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.

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.

$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} y^0$ $+$ $\displaystyle \binom{n}{1} y^1$ $+$ $\displaystyle \binom{n}{2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} y^n$

$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, y^0$ $+$ $^nC_1\, y^1$ $+$ $^nC_2\, y^2$ $+$ $\cdots$ $+$ $^nC_r\, y^r$ $+$ $\cdots$ $+$ $^nC_n\, y^n$

Binomial Theorem in one variable in usual form

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 $x$.

$(1).\,$ $(1+x)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^0$ $+$ $\displaystyle \binom{n}{1} x^1$ $+$ $\displaystyle \binom{n}{2} x^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^n$

$(2).\,$ $(1+x)^n$ $\,=\,$ $^nC_0\, x^0$ $+$ $^nC_1\, x^1$ $+$ $^nC_2\, x^2$ $+$ $\cdots$ $+$ $^nC_r\, x^r$ $+$ $\cdots$ $+$ $^nC_n\, x^n$

Replace the Binomial coefficients by their values

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.

$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}\,x^3$ $+$ $\cdots$