How do you find the sixth term of #(x-1/2)^10#?

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Answer 1 · 2017-06-23T10:08:33

# -63/8x^5.#

We know that, the General #(r+1)^(th)# Term, i.e., #T_(r+1),# in

the expansion of #(a+b)^n# is given by,

#T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.#

We have, #a=x,b=-1/2, n=10, &,# since #T_6# is required, #r=5.#

Accordingly, #T_(5+1)=""_10C_5x^(10-5)(-1/2)^5,#

#=((10)(9)(8)(7)(6))/((1)(2)(3)(4)(5))x^5(-1/2^5),#

# rArr T_6=-63/8x^5.#

Enjoy Maths.!