JEE Main 27 June 2022 Shift 2 Solved Paper

Given, Binomial expression is
=(xn+‌

2

x5

)7
∴ General term
Tr+1=‌7Cr⋅(xn)7−r⋅(‌

2

x5

)r
=‌7Cr⋅x7n−nr−5r⋅2r
For positive power of x,
7n−nr−5r>0
⇒7n>r(n+5)
⇒r<‌

7n

n+5

As r represent term of binomial expression so r is always integer.
Given that sum of coefficient is 939.
When r=0,
sum of coefficient =‌7C0⋅20=1
when r=1,
sum of coefficient =‌7C0⋅20+‌7C1⋅21=1+14=15
when r=2,
sum of coefficient
=‌7C0⋅20+‌7C1⋅21+‌7C2⋅22
=1+14+84
=99
when r=3,
sum of coefficient
=‌7C0⋅20+‌7C1⋅21+‌7C2⋅22+‌7C3⋅23
=1+14+84+280
=379
when r=4,
sum of coefficient
=‌7C0⋅20+‌7C1⋅21+‌7C2⋅22+‌7C3⋅23+‌7C4⋅24
=1+14+84+280+560
=939
To get value of r=4, value of ‌

7n

n+5

should be between 4 and 5 .
∴4<‌

7n

n+5

<5
⇒4n+20<7n<5n+25
∴4n+20<7n
⇒3n>20
⇒n>‌

20

3

⇒n>6.66
and
7n<5n+25
⇒2n<25
⇒n<12.5
∴6.66<n<12.5
∴ Possible integer values of n=7,8,9,10,11,12
∴ Sum of values of n=7+8+9+10+11+12 =57