The value of $ 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $ is

Q: The value of $\large 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $ is

(A) 0

(B) – 1

(C) – 2

(D) – 3

Sol: $\large 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $

$\large = 99^{50} – 99_{C_1}(98)^{50} + 99_{C_2} (97)^{50} – ……+ 99_{C_{98}}.1 $

$\large = 99_{C_0} 99^{50} – 99_{C_1}(99-1)^{50} + 99_{C_2} (99-2)^{50} – ……+ 99_{C_{98}} (99-98)^{50} -99_{C_{99}} (99-99)^{50}$

$\large = 99^{50}( 99_{C_0} – 99_{C_1} + 99_{C_2} – …..+ 99_{C_{98}} -99_{C_{99}} ) $

$\large + 50_{C_1} 99^{49}( 99_{C_1} – 2 . 99_{C_2} + 3 . 99_{C_3} – …… )+ … $

= 0 + 0 +… = 0

Hence (A) is the correct answer.