Question

20. The sum of the coefficients of the first three terms of the expansion of *(x-3/x2) m
is 559. Find the term of the expansion containingx3​

Answer 1

1. Answer

mC0(-3)0+.mC1(-3)1+.mC2(-3)2=559

1+m(-3)+m(m-1)2⋅9=559

-3m+9m2-9m2=558

3m22-5m2=186

3m2-5m-372=0

m=5±√25+12⋅3726

=5+√4464+256

=5+676726=12

to find the term for x3

so,m-3r=3

r=3

so fourth term will be taken t4=.12C3(-3)x3

=12⋅11⋅103⋅2⋅(-27x3)

=-5940x3