Binomial Theorem And Its Simple Applications Question 79
Question: If a and d are two complex numbers, then the sum to $ (n+1) $ terms of the following series $ aC_0-(a+d)C_1+(a+2d)C_2-…….. $ is
Options:
A) $ \frac{a}{2^{n}} $
B) $ na $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- We can write $ aC_0-(a+d),C_1+(a+2d)C_2-…. $ upto $ (n+1) $ terms $ =a(C_0-C_1+C_2-….)+d(-C_1+2C_2-3C_3+….) $ ….(i) Again, $ {{(1-x)}^{n}}=C_0-C_1x+C_2x^{2}-….+{{(-1)}^{n}}C_{n}x^{n} $ …(ii) Differentiating with respect to x, $ -n{{(1-x)}^{n-1}}=-C_1+2C_2x-….+{{(-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+2C_2-….+{{(-1)}^{n}}n.C_{n}=0 $ Thus the required sum to (n+1) terms, by (i) =a.0 + d.0 = 0.
BETA
