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Question : 26 of 100
Marks:
+1,
-0
Solution:
‌(ax−‌)13‌‌ We have, ‌‌Tr+1=‌nCr(p)n−r(q)r‌Tr+1=‌13Cr(ax)13−r(−‌)r‌=‌13Cr(a)13−r(−‌)r(x)13−r⋅(x)−2r‌=‌13Cr(a)13−r(−‌)r(x)13−3r . . . (1)
‌‌ Coefficient of ‌x7‌⇒13−3r=7‌r=2‌r‌ in equation (1) ‌‌T3=‌13C2(a)13−2(−‌)2(x)13−6‌=‌13C2(a)11(‌)2(x)7Coefficient of
x7 is
‌13C2‌Now,
(ax+‌)13‌Tr+1=‌13Cr(ax)13−r(‌)r‌=‌13Cr(a)13−r(‌)r(x)13−r(x)−2r‌=‌13Cr(a)13−r(‌)r(x)13−3r . . . (2)
Coefficient of
x−5‌⇒13−3r=−5‌r=6r in equation
‌T7=‌13C6(a)13−6(‌)6(x)13−18‌T7=‌13C6(a)7(‌)6(x)−5Coefficient of
x−5 is
‌13C6(a)7(‌)6ATQ
Coefficient of
x7= coefficient of
x−5T3=T7‌‌13C2(‌)=‌13C6(a)7(‌)6‌a4⋅b4=‌‌=‌| 13×12×11×10×9×8×1 |
| 13×12×6×5×4×3 |
=22
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