JEE Advanced 2020 Full Test 7 Paper 2

I−P,R‌II−P,R;III−Q;IV−Q,R,S,T
(I) Let consecutive coefficient be ‌nCr and ‌nCr+1. Then ,‌

n!

(n−r)!r!

=‌

n!

(n−r−1)!(r+1)!

⇒‌

1

(n−r)(n−r−1)!r!

=‌

1

(n−r−1)!(r+1)r!

⇒r1=n−r
⇒n=2r+1 hence ,n is odd.
(II) E=(19−4)n+(19+4)n
2[‌nC019n+‌nC219n−2⋅42+............+‌nCn−119.4n−1] when n is even or
2[‌nC019n+‌nC2⋅19n−2⋅42+.........+‌nCn−119.4n−1] when n is odd
⇒E is divisible by 19 when n is odd
(III) ‌10C0‌20C10−‌10C1‌18C10+‌16C2‌16C10−.........
= Coefficient of x10 in [‌10C0(1+x)20−‌20C1×(1+x)18+‌10C2(1+x)16−......]
= Coefficient of x10 in [‌10C0((1+x)2)10−‌10C1×((1+x)2)9+‌10C2((1+x)2)8−........]
= Coefficient of x10 in [(1+x)2−1]10
= Coefficient of x10 in [2x+x2]10
=210
(IV)Tr=‌14Crr−1xr−1;Tr+1=‌14CrXr;Tr+2=‌14Cr+1xr+1
By the given condition, 214Cr=‌14Cr−1+‌14Cr+1
⇒2=‌

‌14Cr−1

‌14Cr

+‌

‌14Cr+1

‌14Cr

⇒2=‌

r

14−r+1

+‌

14−(r+1)+1

r+1

⇒2=‌

r

15−r

+‌

14−r

r+1

⇒r=9