We know that if n C 0 , n C 1 , n C 2 ,……., n C n be binomial coefficients then (1+ x) n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + …..+ C n x n . Various relations among binomial coefficients can be derived by putting x = 1, –1, x = i, x = w
Some other identities can be derived by adding and subtracting two such identities. The expression (a + ib) n can be evaluated by using De-Moiver’s theorem by putting a = r cos θ , b = r sin θ .
(i) The value of n C 0 – n C 2 + n C 4 – n C 6 + ………. must be
2 2n 2 n/2 cos 
2 n 2 n/2 sin 
2 n + 2 n/2 cos 
(ii) The value of the expression ( n C 0 – n C 2 + n C 4 – n C 6 + …….) 2 + ( n C 1 – n C 3 + n C 5 ……) 2 must be None of these (iii) The value of n C 0 + n C 4 + n C 8 +………. must be 2 n–2 + 
cos 
Text Solution
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Ans.
(i)
Sol. Explanation:
For n = 4, the series = 4 C 0 – 4 C 2 + 4 C 4 = 1 – 6 + 1 = – 4
Observe the table
Choice Value of n = 2
4 0 – 4 0
⇒ choice is correct.
Alternative Solution :
(1+ i) n = C 0 + C 1 i + C 2 i 2 + C 3 i 3 + ……
= (C 0 – C 2 + C 4 – C 6 + …..+ i(C 1 –C 3 + C 5 – ……..) (*)
⇒ C 0 – C 2 + C 4 – C 6 +….. = Real part of (1 + i) n = Real part of 
= Real part of 2
n/2 
= 2 n/2 cos 
(ii)
Sol. Explanation:
From the step (*) of alternative solution of above question, it follows that (on taking modulus)
(C 0 – C 2 + C 4 – ……) 2 + (C 1 – C 3 + C 5 – ……) 2
= 2 n
(iii)
Sol. Explanation
We can easily catch the answer by putting n = 4
Alternatively, add the standard identity
C 0 + C 2 + C 4 + C 6 + ….= 2
n–1
into the combinatorial identity
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