ai(1+x)i =a0+a1(1+x)1+a2(1+x)2+a3(1+x)3 +a4(1+x)4+a5(1+x)5 ⇒1+x4+x5 =a0+a1(1+x)+a2(1+2x+x2)+a3(1+3x+3x2+x3) +a4(1+4x+6x2+4x3+x4)+a5(1+5x+10x2+10x3+5x4+x5) ⇒1+x4+x5 =a0+a1+a1x+a2+2a2x+a2x2+a3+3a3x +3a3x2+a3x3+a4+4a4x+6a4x2+4a4x3+a4x4+a5 +5a5x+10a5x2+10a5x3+5a5x4+a5x5 ⇒1+x4+x5 =(a0+a1+a2+a3+a4+a5)+x(a1+2a2+3a3+4a4+5a5) +x2(a2+3a3+6a4+10a5)+x3(a3+4a4+10a5) +x4(a4+5a5)+x5(a5) On comparing the like coefficients, we get a5=1...(i) ; a4+5a5=1 .......(ii) ; a3+4a4+10a5=0 .......(iii) and a2+3a3+6a4+10a5=0...(iv) from (i) & (ii), we get a4=−4...(v) from (i), (iii) & (v), we get a3=+6 ....... (vi) Now, from (i), (v) and (vi), we get a2=−4