Samyak Jain
Last Activity: 5 Years ago
To find : coefficient of x–26 in expansion of (x2 – 2/x4)11.
General term in eapansion of (a + b)n is Tr+1 = nCr . a(n – r). br
Here general term is Tr+1 = 11Cr . (x2)(11 – r). (– 2/x4)r = 11Cr . x2(11 – r). (– 2)r / x4r
i.e. Tr+1 = 11Cr . x22 – 2r – 4r. (– 2)r = 11Cr . x(22 – 6r). (– 2)r
For x
– 26,
x(22 – 6r) = x– 26 i.e. 22 – 6r = – 26 6r = 48or r = 8. Now put r = 8 in Tr+1. We get
T8+1 = 11C8 . x– 26. (– 2)8 = (11 ! / 8 ! 3!) . 28 x– 26 = (11.10.9 / 3.2) . 28 x– 26
T9 = 165 . 28 x– 26 = 330 . 27 x– 26
required coefficient of x– 26 is 330 . 27 .