JEE Main 2022 — Binomial Theorem Question with Solution

Given, Binomial expression is

$$={\left(x^n+\frac{2}{x^5}\right)}^7$$

$$\therefore$$ General term

$$T_{r+1}={}^7{C}_r.(x^n{)}^{7-r}.{\left(\frac{2}{x^5}\right)}^r$$

$$={}^7{C}_r.x^{7n-nr-5r}.2^r$$

For positive power of x,

$$7n-nr-5r>0$$

$$\Rightarrow 7n>r(n+5)$$

$$\Rightarrow r<\frac{7n}{n+5}$$

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 $$={}^7{C}_0.2^0=1$$

when $$r=1$$,

sum of coefficient $$={}^7{C}_0.2^0+{}^7{C}_1.2^1=1+14=15$$

when $$r=2$$,

sum of coefficient

$$={}^7{C}_0.2^0+{}^7{C}_1.2^1+{}^7{C}_2.2^2$$

$$=1+14+84$$

$$=99$$

when $$r=3$$,

sum of coefficient

$$={}^7{C}_0.2^0+{}^7{C}_1.2^1+{}^7{C}_2.2^2+{}^7{C}_3.2^3$$

$$=1+14+84+280$$

$$=379$$

when $$r=4$$,

sum of coefficient

$$={}^7{C}_0.2^0+{}^7{C}_1.2^1+{}^7{C}_2.2^2+{}^7{C}_3.2^3+{}^7{C}_4.2^4$$

$$=1+14+84+280+560$$

$$=939$$

$$\therefore$$ For r = 4 sum of coefficient = 939

To get value of r = 4, value of $$\frac{7n}{n+5}$$ should be between 4 and 5.

$$\therefore$$ $$4<\frac{7n}{n+5}<5$$

$$\Rightarrow 4n+20<7n<5n+25$$

$$\therefore$$ $$4n+20<7n$$

$$\Rightarrow 3n>20$$

$$\Rightarrow n>\frac{20}{3}$$

$$\Rightarrow n>6.66$$

and

$$7n<5n+25$$

$$\Rightarrow 2n<25$$

$$\Rightarrow n<12.5$$

$$\therefore$$ $$6.66<n<12.5$$

$$\therefore$$ Possible integer values of $$n=7,8,9,10,11,12$$

$$\therefore$$ Sum of values of $$n=7+8+9+10+11+12$$

$$=57$$