Difference between revisions of "Binomial theorem"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| Line 5: | Line 6: | ||
==Proof== | ==Proof== | ||
| + | We will proceed using [[induction]]. If $n=0$ then | ||
| + | $$(a+b)^0 = 1 = {0 \choose 0} a^0 b^0 = \displaystyle\sum_{k=0}^0 {0 \choose k} a^k b^{0-k}.$$ | ||
| + | We assume that | ||
| + | $$(a+b)^n = \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k}.$$ | ||
| + | Now compute using [[binomial coefficient (n choose 0) equals 1]] and [[binomial coefficient ((n+1) choose k) equals (n choose k) + (n choose (k-1))]], we see | ||
| + | $$\begin{array}{ll} | ||
| + | (a+b)^{n+1} &= (a+b) (a+b)^n \\ | ||
| + | &=(a+b) \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k} \\ | ||
| + | &= \left[ \displaystyle\sum_{k=0}^n {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k+1} \right] \\ | ||
| + | &= \left[ \displaystyle{n \choose n} a^{n+1} + \displaystyle\sum_{k=0}^{n-1} {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle{n \choose 0} b^{n+1}+ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ | ||
| + | &=a^{n+1} + b^{n+1} + \left[ \displaystyle\sum_{k=0}^{n-1} {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ | ||
| + | &=a^{n+1} + b^{n+1} + \left[ \displaystyle\sum_{k=1}^{n} {n \choose {k-1}} a^{k} b^{n-k+1} \right] + \left[ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ | ||
| + | &= \displaystyle{{n+1} \choose 0} a^{n+1} + \displaystyle{{n+1} \choose {n+1}} b^{n+1} + \displaystyle\sum_{k=1}^n \left[ {n \choose {k-1}} + {n \choose k} \right] a^k b^{n-k+1} \\ | ||
| + | &=\displaystyle{{n+1} \choose 0} a^{n+1} + \displaystyle{{n+1} \choose {n+1}} b^{n+1} + \displaystyle\sum_{k=1}^n {{n+1} \choose k} a^k b^{n-k+1} \\ | ||
| + | &= \displaystyle\sum_{k=0}^{n+1} {{n+1} \choose k} a^k b^{n-k+1}, | ||
| + | \end{array}$$ | ||
| + | proving the claim. | ||
| + | |||
| + | ==External links== | ||
| + | [https://proofwiki.org/wiki/Binomial_Theorem/Integral_Index Proofwiki] | ||
==References== | ==References== | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|next=Binomial coefficient}}: $3.1.1$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 18:09, 25 September 2016
Theorem
The following formula holds: $$(a+b)^n = \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k},$$ where ${n \choose k}$ denotes the binomial coefficient
Proof
We will proceed using induction. If $n=0$ then $$(a+b)^0 = 1 = {0 \choose 0} a^0 b^0 = \displaystyle\sum_{k=0}^0 {0 \choose k} a^k b^{0-k}.$$ We assume that $$(a+b)^n = \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k}.$$ Now compute using binomial coefficient (n choose 0) equals 1 and binomial coefficient ((n+1) choose k) equals (n choose k) + (n choose (k-1)), we see $$\begin{array}{ll} (a+b)^{n+1} &= (a+b) (a+b)^n \\ &=(a+b) \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k} \\ &= \left[ \displaystyle\sum_{k=0}^n {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=0}^n {n \choose k} a^k b^{n-k+1} \right] \\ &= \left[ \displaystyle{n \choose n} a^{n+1} + \displaystyle\sum_{k=0}^{n-1} {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle{n \choose 0} b^{n+1}+ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ &=a^{n+1} + b^{n+1} + \left[ \displaystyle\sum_{k=0}^{n-1} {n \choose k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ &=a^{n+1} + b^{n+1} + \left[ \displaystyle\sum_{k=1}^{n} {n \choose {k-1}} a^{k} b^{n-k+1} \right] + \left[ \displaystyle\sum_{k=1}^{n} {n \choose k} a^k b^{n-k+1} \right] \\ &= \displaystyle{{n+1} \choose 0} a^{n+1} + \displaystyle{{n+1} \choose {n+1}} b^{n+1} + \displaystyle\sum_{k=1}^n \left[ {n \choose {k-1}} + {n \choose k} \right] a^k b^{n-k+1} \\ &=\displaystyle{{n+1} \choose 0} a^{n+1} + \displaystyle{{n+1} \choose {n+1}} b^{n+1} + \displaystyle\sum_{k=1}^n {{n+1} \choose k} a^k b^{n-k+1} \\ &= \displaystyle\sum_{k=0}^{n+1} {{n+1} \choose k} a^k b^{n-k+1}, \end{array}$$ proving the claim.
External links
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (next): $3.1.1$
