For a positive integer $n$: $(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r$ There are $n+1$ terms

Question

For a positive integer $n$: $(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r$ There are $n+1$ terms. Powers of $a$ decrease as powers of $b$ increase; their sum is $n$. Binomial coefficients $\binom nr=\dfrac{n!}{r!(n-r)!}\qquad r=0,1,\ldots,n$ $0!=1\qquad\binom n0=\binom nn=1\qquad\binom n1=n\qquad\binom nr=\binom n{n-r}$ CoefficientEvaluation $\binom{10}{2}$$\dfrac{10\times9}{2!}=45$ $\binom73$$\dfrac{7\times6\times5}{3!}=35$ Use the calculator $n\mathrm Cr$ function for larger coefficients.

Accepted Answer

$(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r$, where $\binom nr=\frac{n!}{r!(n-r)!}$ and $T_{r+1}=\binom nr a^{n-r}b^r$

Coefficient	Evaluation
\(\binom{10}{2}\)	\(\dfrac{10\times9}{2!}=45\)
\(\binom73\)	\(\dfrac{7\times6\times5}{3!}=35\)

Binomial Theorem

Binomial Theorem

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