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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 $(a-b)^n$ ?

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  • $\begingroup$ Try letting $x = -b$. $\endgroup$
    – Burrrrb
    Jul 16 '16 at 11:11
  • $\begingroup$ Yes, just substitute for b for -b $\endgroup$ Jul 16 '16 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..... $\endgroup$
    – Soham
    Jul 16 '16 at 12:54
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Hint:$$\\ (a+b)^{ n }=\sum _{ k=0 }^{ n }{ \binom{n}{k} { a }^{ n-k } } { b }^{ k }$$ now substitute here $b=-b$ $$(a-b)^{ n }=\sum _{ k=0 }^{ n }{ { \left( -1 \right) }^{ k }\binom{n}{k} { a }^{ n-k } } { b }^{ k }$$

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    $\begingroup$ 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 R
    Jul 16 '16 at 13:05
  • $\begingroup$ @WillR,thank you,but think it is quite clear what i've meant $\endgroup$
    – haqnatural
    Jul 16 '16 at 13:12
  • $\begingroup$ maybe $b\mapsto -b$ would be more clear notation $\endgroup$
    – C Squared
    Dec 10 '20 at 14:16
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$$(a-b)^n = (a+ (-b))^n = \sum_{k=0}^n \binom{n}{k}a^k (-b)^{n-k} = \sum_{k=0}^n \binom{n}{k}(-1)^{n-k}a^kb^{n-k} = $$

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