Question

Coefficient of x⁴⁹ in the expansion of (x–1)(x–3)(x–5).........................(x–99) is 

(a)  –99² 

(b)  1 

(c)  –2500

(d)  None of these 

Answer 1

Correct option (c) –2500 

Explanation:

(x–1)(x–3)(x–5) ................ (x–99)

= x⁵⁰ – S₁ x⁴⁹ + S₂ x ⁴⁸ .............

Coefficient of x⁴⁹ is – S₁

= –(1+3+5+......+99) 

= –50² = – 2500 

Comments

How to find S2 and S3 in this case please tell me

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S_2 = coefficient of x^48 = (1.3+1.5+1.7+...+1.99)+(3.5+3.7+3.9+...+3.99)+(5.7+5.9+...+5.99)+...+(95.97+95.99)+97.99, similarly, we can find coefficient of x^47 which is equals to S_3