find the 13th term of the binomial expansion of: (2x-4)^21

Question

Algebra 2 class work please help!

 

Answer 1

The Binomial expansion of (1 + x ) ⁿ is,

1 + n x + [ n(n-1).x ² ] / 2 + [ n(n-1)(n-2).x ³ ] / 3! + .... ..

with the k th term given as tk = n(n-1)(n-2) ... (n - (k-2)).x ^ (k-1)/(k-1)!, k > 2

We have (2 x - 4) ²¹, which becomes

(- 4) ²¹[1 - ( x / 2)] ²¹

(- 4) ²¹[1 + ( - x / 2)] ²¹

The expansion of this is,

(- 4) ²¹{1 + 21(- x / 2) + [ 21*20*(- x / 2) ² ] / 2 +[ 21*20*19*(- x /2)³  ] /3! + ...

And the 13th term (k=13) is,

tk =[ n(n-1)(n-2) ... (n - (k-2))* x ^(k-1)] / (k-1)! , k > 2

t13 = ( -4) ²¹ [( 21*20*19* .... *12*11*10 )*(- x /2) ¹² ] / 12!

=[ (- 4) ²¹*146,965 * x ¹² * 2 ^-12] / 2048

= [ -  4 ^{6 *}  4¹⁵ * 146,965 * x ¹² * 2 ^-12  ] / 2048

= [ - 2 ¹² * 4¹⁵ * 146,965 x ¹²  * 2 ^-12 ] / 2048

= [ -  4¹⁵ * 146,965 x ¹²   ] / 2048

t13 = - 293,930 * 4¹⁵ *  x ¹². 

Solution : 13th term in the expansion is - 293,930 * 4¹⁵ *  x ¹² .