JEE Main 2023 — Mathematics Algebra

We know that ${(1-x)}^{100}=C0100-C1100\cdot x+C2100\cdot {x}^{2}-.......-C49100\cdot {x}^{49}+.......+C100100\cdot {x}^{100}$(1−x)100=C0100−C1100⋅x+C2100⋅x2−.......−C49100⋅x49+.......+C100100⋅x100

Now let, the sum of first fifty coefficients be $S$S.

$$\Rightarrow S=C0100-C1100+C2100-.......-C49100$$⇒S=C0100−C1100+C2100−.......−C49100

Let us substitute $x=1$x=1 in the expansion.

$$\Rightarrow {(1-1)}^{100}=C0100-C1100\cdot 1+C2100\cdot {1}^{2}-.......-C49100\cdot {1}^{49}+.......+C100100\cdot {1}^{100}$$⇒(1−1)100=C0100−C1100⋅1+C2100⋅12−.......−C49100⋅149+.......+C100100⋅1100

$$\Rightarrow 0=C0100-C1100+C2100-.......-C49100+C50100-C51100+C52100.......+C100100$$⇒0=C0100−C1100+C2100−.......−C49100+C50100−C51100+C52100.......+C100100

There are total of $101$101 terms and the middle term is ${T}_{51}=C50100$T51​=C50100.

As $Crn=Cn-rn$Crn=Cn−rn

$$\Rightarrow 0=C0100-C1100+C2100-.......-C49100+C50100-C49100+C48100.......+C0100$$⇒0=C0100−C1100+C2100−.......−C49100+C50100−C49100+C48100.......+C0100

$$\Rightarrow 0=C50100+C0100-C1100+C2100-.......-C49100-C49100+C48100.......+C0100$$⇒0=C50100+C0100−C1100+C2100−.......−C49100−C49100+C48100.......+C0100

$$\Rightarrow 0=C50100+2(C0100-C1100+C2100-.......-C49100)$$⇒0=C50100+2(C0100−C1100+C2100−.......−C49100)

$$\Rightarrow -\frac{C50100}{2}=(C0100-C1100+C2100-.......-C49100)$$⇒−2C50100​=(C0100−C1100+C2100−.......−C49100)

$$\Rightarrow C0100-C1100+C2100-.......-C49100=-\frac{C50100}{2}$$⇒C0100−C1100+C2100−.......−C49100=−2C50100​

Therefore, the required answer is $-\frac{C50100}{2}=-\frac{1}{2}\times \frac{100}{50}C4999=-C4999$−2C50100​=−21​×50100​C4999=−C4999.