Asked by Rich
find coefficient of x^8 in expanding of (x-2)^11
Answers
Answered by
Reiny
the general term t(n+1) = C(11,n) x^(11-n) (-1)^n
so 11-n = 8
n = 3
so the 4th term is
C(11,3) x^8 (-1)^3
= -165x^8
or ..... start expanding using the elements of Pascals triangle
http://ptri1.tripod.com/
(x-2)^11
= <b>1</b>x^11 (-1)^0 + <b>11</b>x^10 (-1)^1 + <b>55</b>x^9 (-1)^2 + <b>165</b>x^8 (-1)^3 + ...
= x^11 - 11x^10 + 55x^9 <b>- 165x^8</b> + 330x^7 - ...
so 11-n = 8
n = 3
so the 4th term is
C(11,3) x^8 (-1)^3
= -165x^8
or ..... start expanding using the elements of Pascals triangle
http://ptri1.tripod.com/
(x-2)^11
= <b>1</b>x^11 (-1)^0 + <b>11</b>x^10 (-1)^1 + <b>55</b>x^9 (-1)^2 + <b>165</b>x^8 (-1)^3 + ...
= x^11 - 11x^10 + 55x^9 <b>- 165x^8</b> + 330x^7 - ...
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