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Question

Mathematics Question on Binomial theorem

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

A

78\frac {7} {8}

B

87\frac {8} {7}

C

2122\frac {21} {22}

D

2324\frac {23} {24}

Answer

78\frac {7} {8}

Explanation

Solution

T21=T20+1= 44c20x20T_{21} = T_{20+1} = \,^{44}c_{20}x^{20} T22=T21+1= 44c21(−x)21=− 44c21x21T_{22} = T_{21+1} = \,^{44}c_{21}\left(-x\right)^{21} = - \,^{44}c_{21}x^{21} ∴ 44c20x20=− 44c21x21\therefore \,^{44}c_{20}x^{20} = - \,^{44}c_{21}x^{21} ∴x=44c2044c21=44 !24 ! 20 !×21 ! 3!44 !\therefore x = \frac{^{44}c_{20}}{^{44}c_{21}} = \frac{44\,!}{24\,!\,20\,!} \times \frac{21\,!\,3!}{44\,!} =21⋅20 ! 23 !20 ! 24⋅23 !=−2124=−78= \frac{21\cdot20\,!\,23\,!}{20\,!\,24\cdot23\,!} = -\frac{21}{24} = -\frac{7}{8}