Find the 24th term in the expansion of #(a+b)^25#?

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Answer 1 · 2018-03-27T16:50:28

#t_24=-300 *a^2 *b^23#

We know that , in the expansion of #(a+b)^n#

#T_(r+1)=(-1)^rC_r^n(a)^(n-r)(b)^r#

Comparing #(a+b)^25 with, (a+b)^n#

#n=25 and r+1=24=>r=23#

So,

#T_(23+1)=(-1)^23 C_23^25 (a)^(25-23)(b)^23....to(I)#

Now ,#(-1)^23=-1#

and #C_r^n=C_(n-r)^n#

#=>C_23^25= C_(25-23)^25=C_2^25=(25xx24)/(2xx1)=300#

Hence, from #(I)#

#t_24=-300 *a^2 *b^23#