UPSC NDA I 2021 Mathematics Solved Paper

Concept
(1+x)n=‌nC0×1(n−0)×x0+‌nC1×1(n−1)×x1+‌nC2×1(n−2)×x2+...++‌nCn×1(n−n)×xn
n‌th ‌ term of the G.P. is an =arn−1
Sum of n terms =s=‌

a(rn−1)

(r−1)

‌

a(rn−1)

(r−1)

; where r>1
Calculation
Sum of n terms =s=‌

a(1−rn)

(1−r)

‌

a(1−rn)

(1−r)

; where r<1
C(n,1)+C(n,2)+−−−−+C(n,n)
‌‌=‌nC1+‌nC2+...+‌nCn
‌‌=‌nC0+‌nC1+‌nC2+...+‌nCn−‌nC0
‌‌=(1+1)n−‌nC0
‌‌=2n−1=‌

2n−1

2−1

‌

2n−1

2−1

=1×‌

2n−1

2−1

‌

2n−1

2−1

Comparing it with a G.P sum =a×‌

rn−1

r−1

‌

rn−1

r−1

, we get a=1 and r=2
So, 2n−1=1+2+22+...+2n−1 which will give us n terms in total.