(1+x)21+(1+x)22+..+(1+x)30 in the expansion of this what is the coefficient of x5 is
হানি নাটস
Let
S=(1+x)21+(1+x)22+(1+x)23+…(1+x)30…(i)(1+x)S=(1+x)22+(1+x)23…(1+x)30+(1+x)31
Subtracting (i) from (ii), we get
xS=(1+x)31−(1+x)21
S=x(1+x)31−x(1+x)21
Coefficient of xr
=31Crxr−1−21Crxr−1=(31Cr−21Cr)xr−1…(a)
Therefore the coefficient of x5 implies
r−1=5
r=6
Substituting in (a), we get a coefficient of x5
=(31C6−21C6)