- Home
- Standard 11
- Mathematics
7.Binomial Theorem
normal
Find the coefficient of $x^{49}$ in the expansion of $(2x + 1) (2x + 3) (2x + 5)----- (2x + 99)$
A
${2^{50}} \times 2500$
B
${2^{49}} \times 2500$
C
${-2^{50}} \times 2500$
D
${-2^{49}} \times 2500$
Solution
$(2 n+1)(2 n+3)(2 n+5) \dots(2 x+99)$
$\Rightarrow 2^{50}\left[\left(\mathrm{x}+\frac{1}{2}\right)\left(\mathrm{x}+\frac{3}{2}\right)\left(\mathrm{x}+\frac{5}{2}\right) \ldots\left(\mathrm{x}+\frac{99}{2}\right)\right]$
Coefficient of $\mathrm{x}^{49}$ is equal to $-(\mathrm{sum} \text { of roots })$
$=2^{50}\left[\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\ldots .+\frac{99}{2}\right]=2^{49} \times 2500$
Standard 11
Mathematics
