A) \[\frac{a}{{{2}^{n}}}\]
B) \[na\]
C) 0
D) None of these
Correct Answer: C
Solution :
We can write \[a{{C}_{0}}-(a+d)\,{{C}_{1}}+(a+2d){{C}_{2}}-....\]upto \[(n+1)\]terms \[=a({{C}_{0}}-{{C}_{1}}+{{C}_{2}}-....)+d(-{{C}_{1}}+2{{C}_{2}}-3{{C}_{3}}+....)\] ....(i) Again,\[{{(1-x)}^{n}}={{C}_{0}}-{{C}_{1}}x+{{C}_{2}}{{x}^{2}}-....+{{(-1)}^{n}}{{C}_{n}}{{x}^{n}}\] ...(ii) Differentiating with respect to x, \[-n{{(1-x)}^{n-1}}=-{{C}_{1}}+2{{C}_{2}}x-....+{{(-1)}^{n}}{{C}_{n}}n{{x}^{n-1}}\] ....(iii) Putting x =1 in (ii) and (iii), we get \[{{C}_{0}}-{{C}_{1}}+{{C}_{2}}-....+{{(-1)}^{n}}{{C}_{n}}=0\] and \[-{{C}_{1}}+2{{C}_{2}}-....+{{(-1)}^{n}}n.{{C}_{n}}=0\] Thus the required sum to (n+1) terms, by (i) =a.0 + d.0 = 0.
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