Is it true that #(a+b)^40 = sum_(k=0)^40 ((40),(k)) a^(40-k) b^k# ?

1 Answer
Nov 26, 2017

The binomial series works for any positive integer power, so yes, it's right.

Explanation:

The binomial theorem tells us that:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#

where #((n),(k)) = (n!)/((n-k)! k!)#

This holds for any positive integer #n#.

So:

#(a+b)^40 = sum_(k=0)^40 ((40),(k)) a^(40-k) b^k#

#color(white)((a+b)^40) = a^40 + 40 a^39 b + 780 a^38 b^2 + 9880 a^37 b^3 + 91390 a^36 b^4 + 658008 a^35 b^5 + 3838380 a^34 b^6 + 18643560 a^33 b^7 + 76904685 a^32 b^8 + 273438880 a^31 b^9 + 847660528 a^30 b^10 + 2311801440 a^29 b^11 + 5586853480 a^28 b^12 + 12033222880 a^27 b^13 + 23206929840 a^26 b^14 + 40225345056 a^25 b^15 + 62852101650 a^24 b^16 + 88732378800 a^23 b^17 + 113380261800 a^22 b^18 + 131282408400 a^21 b^19 + 137846528820 a^20 b^20 + 131282408400 a^19 b^21 + 113380261800 a^18 b^22 + 88732378800 a^17 b^23 + 62852101650 a^16 b^24 + 40225345056 a^15 b^25 + 23206929840 a^14 b^26 + 12033222880 a^13 b^27 + 5586853480 a^12 b^28 + 2311801440 a^11 b^29 + 847660528 a^10 b^30 + 273438880 a^9 b^31 + 76904685 a^8 b^32 + 18643560 a^7 b^33 + 3838380 a^6 b^34 + 658008 a^5 b^35 + 91390 a^4 b^36 + 9880 a^3 b^37 + 780 a^2 b^38 + 40 a b^39 + b^40#