Question 1034947
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Find the 6th term of (-4a+3b)^13
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  1-st term               2-nd term              3-rd term             4-th term             5-th term            6-th term

{{{C[13]^0*(-4a)^13*(3b)^0}}}   {{{C[13]^1*(-4a)^12*(3b)^1}}}   {{{C[13]^2*(-4a)^11*(3b)^2}}}   {{{C[13]^3*(-4a)^10*(3b)^3}}}   {{{C[13]^4*(-4a)^9*(3b)^4}}}   {{{C[13]^5*(-4a)^8*(3b)^5}}}


The 6-th term is {{{C[13]^5*(-4a)^8*(3b)^5}}}  (the general formula is  {{{C[n]^k*(-4a)^(n-k)*(3b)^k}}},  k = 0 (first term), 1 (the 2-nd term), 2 (the 3-rd term), . . . ).

Or, after simplifying, {{{C[13]^5*(-4a)^8*(3b)^5}}} = {{{C[13]^5*4^8*3^5*a^8*b^5}}} = {{{1287*4^8*3^5*a^8*b^5}}} = {{{20495794176*a^8*b^5}}}.


{{{C[13]^5}}} = {{{(13!)/(5!*8!)}}} = {{{(13*12*11*10*9)/(1*2*3*4*5)}}} = 1287  is the number of combination 13 things taken 5 at a time.


On combinations, see the lesson <A HREF=https://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A> in this site.


On binomial expansion see the lesson <A HREF=https://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>.
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