If the sum of the coefficients of all the positive powers of x , in the binomial expansion of ( x ^ n + 2 x ^ 5 ) ^ 7 is 939 , then the sum of all the possible integral values of n is

Correct answer: 57 Plotting the table for coefficient of ( x ^ n + 2 x ^ 5 ) ^ 7 and there cumulative sum : Coefficient Commulative sum x ^ 7n → C0 7 1 x ^ 6n - 5 → 2 × C1 7 1 + 14 x ^ 5n - 10 → 2 ^ 2 × C2 7 1 + 14 + 84 x ^ 4n - 15 → 2 ^ 3 × C3 7 1 + 14 + 84 + 280 x ^ 3n - 20 → 2 ^ 4 × C4 7 1 + 4 + 84 + 280 + 560 = 939 x ^ 2n - 25 → 2 ^ 5 × C5 7 So, 3n - 20 ≥ 0 2n - 25 ∴ 7 ≤ n ≤ 12 Sum = 7 + 8 + 9 + 10 + 11 + 12 = 57

JEE Main 202227 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2022 · 27 June · Evening Shift · Mathematics · Binomial Theorem

If the sum of the coefficients of all the positive powers of x , in...

If the sum of the coefficients of all the positive powers of

x

, in the binomial expansion of

{\left({x}^{n} + \dfrac{2}{{x}^{5}}\right)}^{7} is

939

, then the sum of all the possible integral values of

n$ is

Correct answer

Step-by-step solution

Plotting the table for coefficient of ${\left({x}^{n} + \dfrac{2}{{x}^{5}}\right)}^{7} and there cumulative sum :</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>Coefficient</td> <td>Commulative sum</td> </tr> <tr> <td>$ {x}^{7n} \to C0 7 $</td> <td>$ 1 $</td> </tr> <tr> <td>$ {x}^{6n - 5} \to 2 \times C1 7 $</td> <td>$ 1 + 14 $</td> </tr> <tr> <td>$ {x}^{5n - 10} \to {2}^{2} \times C2 7 $</td> <td>$ 1 + 14 + 84 $</td> </tr> <tr> <td>$ {x}^{4n - 15} \to {2}^{3} \times C3 7 $</td> <td>$ 1 + 14 + 84 + 280 $</td> </tr> <tr> <td>$ {x}^{3n - 20} \to {2}^{4} \times C4 7 $</td> <td>$ 1 + 4 + 84 + 280 + 560 = 939 $</td> </tr> <tr> <td>$ {x}^{2n - 25} \to {2}^{5} \times C5 7 $</td> <td> </td> </tr> </tbody></table><p>So,$ 3n - 20 \ge 0 \cap 2n - 25 < 0 \cap n \in I $</p><p>$ ∴ 7 \le n \le 12 $</p><p>Sum$ = 7 + 8 + 9 + 10 + 11 + 12 = 57

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