Question #142110
Use Mathematical Induction to prove the Binomial Theorem.
1
Expert's answer
2020-11-17T17:05:26-0500

Let us give a proof of the Binomial Theorem using mathematical induction. We will need to use Pascal's identity in the form:


(nr1)+(nr)=(n+1r)\left(\begin{array}{c}n\\r-1\end{array}\right)+\left(\begin{array}{c}n\\r\end{array}\right)=\left(\begin{array}{c}n+1\\r\end{array}\right) for 0<rn0<r\le n.


We aim to prove that


(a+b)n=an+(n1)an1b+(n2)an2b2++(nr)anrbr++(nn1)abn1+bn.(a+b)^n=a^n+\left(\begin{array}{c}n\\1\end{array}\right)a^{n−1}b+\left(\begin{array}{c}n\\2\end{array}\right)a^{n−2}b^2+⋯+\left(\begin{array}{c}n\\r\end{array}\right)a^{n−r}b^r +⋯+\left(\begin{array}{c}n\\n-1\end{array}\right)ab^{n−1}+ b^n.


We first note that the result is true for n=1 and n=2: (a+b)1=a+b(a+b)^1=a+b and (a+b)2=a2+2ab+b2=a2+(21)ab+b2(a+b)^2=a^2+2ab+b^2=a^2+\left(\begin{array}{c}2\\1\end{array}\right)ab+b^2.


Let kk be a positive integer with k2k≥2 for which the statement is true. So


(a+b)k=ak+(k1)ak1b+(k2)ak2b2++(kr)akrbr++(kk1)abk1+bk(a+b)^k=a^k+\left(\begin{array}{c}k\\1\end{array}\right)a^{k−1}b+\left(\begin{array}{c}k\\2\end{array}\right)a^{k−2}b^2+⋯+\left(\begin{array}{c}k\\r\end{array}\right)a^{k−r}b^r +⋯+\left(\begin{array}{c}k\\k-1\end{array}\right)ab^{k−1}+ b^k


Now consider the expansion


(a+b)k+1=(a+b)(a+b)k=(a+b)^{k+1}=(a+b)(a+b)^k=


=(a+b)(ak+(k1)ak1b+(k2)ak2b2++(kr)akrbr++(kk1)abk1+bk)==(a+b)(a^k+\left(\begin{array}{c}k\\1\end{array}\right)a^{k−1}b+\left(\begin{array}{c}k\\2\end{array}\right)a^{k−2}b^2+⋯+\left(\begin{array}{c}k\\r\end{array}\right)a^{k−r}b^r +⋯+\left(\begin{array}{c}k\\k-1\end{array}\right)ab^{k−1}+ b^k)=


=ak+1+[1+(k1)]akb+[(k1)+(k2)]ak1b2++[(kr1)+(kr)]ak+1rbr++[(kk1)+1]abk+bk+1=a^{k+1}+\left[1+\left(\begin{array}{c}k\\1\end{array}\right)\right]a^{k}b+\left[\left(\begin{array}{c}k\\1\end{array}\right)+\left(\begin{array}{c}k\\2\end{array}\right)\right]a^{k−1}b^2+⋯+\left[\left(\begin{array}{c}k\\r-1\end{array}\right)+\left(\begin{array}{c}k\\r\end{array}\right)\right]a^{k+1−r}b^r +⋯+\left[\left(\begin{array}{c}k\\k-1\end{array}\right)+1\right]ab^{k}+ b^{k+1}


From Pascal's identity, it follows that


(a+b)k+1=ak+1+(k+11)akb+(k+12)ak1b2++(k+1r)ak+1rbr++(k+1k)abk+bk+1(a+b)^{k+1}=a^{k+1}+\left(\begin{array}{c}k+1\\1\end{array}\right)a^{k}b+\left(\begin{array}{c}k+1\\2\end{array}\right)a^{k−1}b^2+⋯+\left(\begin{array}{c}k+1\\r\end{array}\right)a^{k+1−r}b^r +⋯+\left(\begin{array}{c}k+1\\k\end{array}\right)ab^{k}+ b^{k+1}


Hence the result is true for k+1k+1. By induction, the result is true for all positive integers nn.




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