Question 432996
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Hi
In General: (a+b)^n = {{{sum( nCk(a^(n-k))(b^k), k=0, n )}}} where nCk = {{{n!/(x!(n-k)!)}}}
It is relatively easy to understand the expansion in terms of variable's exponents
the coefficients very do-able, once one understands how nCk works with the
denominator canceling out a great deal of the numerator.
5C0 = 5!/0!5! = 1  5C1 = 5!/1!4! = 5  5C2 = 5!/2!3! = 10 etc
(2x+5)^5 =  1a^5 +   5a^4b    + 10a^3b^2     + 10a^2b^3 +     5ab^4 +    1b^5
          5C0       5C1        5C2            5C3            5C4        5C5
(2x +5)^5=  (2x)^5 + 5(2x)^4*5 + 10(2x)^3*25 + 10(2x)^2*125 + 5(2x)*625 + 3125
  Will leave the arithmetic up to Your Calculator

{{{(a + b)^0 = 1}}}
{{{(a + b)^1 = a + b}}}
{{{(a + b)^2 = a^2 + 2ab + b^2 }}}
{{{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 }}}
{{{(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 }}}
{{{(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 }}}