If the 21st and 22nd terms in the expansion of (1 + x)44, are equal, then x =

Question

If the 21st and 22nd terms in the expansion of (1 + x)44, are equal, then x = (A)  (7 /8) (B)  (8 /7) (C)  (21 /22) (D)  (23 /24)

Tardigrade Admin · Accepted Answer

T21 = T20+1 = 44c20x20 T22 = T21+1 = 44c21(-x)21 = - 44c21x21 ∴ 44c20x20 = - 44c21x21 ∴ x = (44c20/44c21) = (44 !/24 ! 20 !) × (21 ! 3!/44 !) = (21⋅20 ! 23 !/20 ! 24⋅23 !) = -(21/24) = -(7/8)

Q. If the 21st and 22nd terms in the expansion of $(1 + x)^{44}$ , are equal, then x =

$\frac{7}{8}$

$\frac{8}{7}$

$\frac{21}{22}$

$\frac{23}{24}$

Q. If the 21st and 22nd terms in the expansion of (1+x)44, are equal, then x =

87​

78​

2221​

2423​

T21​=T20+1​=44c20​x20 T22​=T21+1​=44c21​(−x)21=−44c21​x21 ∴44c20​x20=−44c21​x21 ∴x=44c21​44c20​​=24!20!44!​×44!21!3!​ =20!24⋅23!21⋅20!23!​=−2421​=−87​

Q. If the 21st and 22nd terms in the expansion of $(1 + x)^{44}$ , are equal, then x =

$\frac{7}{8}$

$\frac{8}{7}$

$\frac{21}{22}$

$\frac{23}{24}$