Like there is a formula for the binomial expansion of $(a+b)^n$ that can be neatly and compactly be written as a summation, does there exist an equivalent formula for $(a-b)^n$ ?
$\begingroup$
$\endgroup$
3
asked Jul 16, 2016 at 11:08
-
$\begingroup$ Try letting $x = -b$. $\endgroup$– BurrrrbJul 16, 2016 at 11:11
-
$\begingroup$ Yes, just substitute for b for -b $\endgroup$– samerivertwiceJul 16, 2016 at 12:27
-
$\begingroup$ Ok...so you know the formula for $(a+b)^n$...now,try this...$(a-b)^n=\{a+(-b)\}^n$....Now,you know the formula..... $\endgroup$– SohamJul 16, 2016 at 12:54
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Hint:$$\\ (a+b)^{ n }=\sum _{ k=0 }^{ n }{ \binom{n}{k} { a }^{ n-k } } { b }^{ k }$$ now substitute here $b=-b$ $$(a-b)^{ n }=\sum _{ k=0 }^{ n }{ { \left( -1 \right) }^{ k }\binom{n}{k} { a }^{ n-k } } { b }^{ k }$$
answered Jul 16, 2016 at 11:12
-
2$\begingroup$ This answer could be made clearer. By setting $b=-b$ you are effectively setting $b=0$. While I realise your intent, it could lead to confusion for others. $\endgroup$– Will RJul 16, 2016 at 13:05
-
$\begingroup$ @WillR,thank you,but think it is quite clear what i've meant $\endgroup$ Jul 16, 2016 at 13:12
-
$\begingroup$ maybe $b\mapsto -b$ would be more clear notation $\endgroup$ Dec 10, 2020 at 14:16
$\begingroup$
$\endgroup$
$$(a-b)^n = (a+ (-b))^n = \sum_{k=0}^n \binom{n}{k}a^k (-b)^{n-k} = \sum_{k=0}^n \binom{n}{k}(-1)^{n-k}a^kb^{n-k} = $$
