Q.23. If the sum of the coefficients of and in the expansion of , , is zero, then the value of n i

Q: What is the correct answer for JEE Main Mathematics Q23?

The correct answer is 21. Sol. The general term in the expansion of $(x^2 + \frac{1}{x^3})^n$ is $T_{r+1} = \binom{n}{r} (x^2)^{n-r} (x^{-3})^r = \binom{n}{r} x^{2n-2r-3r} = \binom{n}{r} x^{2n-5r}$. For the coefficient of $x^7$, we set $2n-5r = 7 \Rightarrow 5r = 2n-7$. So $r = \frac{2n-7}{5}$. For the coefficient of $x^{11}$, we set $2n-5r' = 11 \Rightarrow 5r' = 2n-11$. So $r' = \frac{2n-11}{5}$. Since r and r' must be n

Q: Q.23. If the sum of the coefficients of and in the expansion of , , is zero, then the value of n is

The correct answer is 21. Sol. The general term in the expansion of $(x^2 + \frac{1}{x^3})^n$ is $T_{r+1} = \binom{n}{r} (x^2)^{n-r} (x^{-3})^r = \binom{n}{r} x^{2n-2r-3r} = \binom{n}{r} x^{2n-5r}$. For the coefficient of $x^7$, we set $2n-5r = 7 \Rightarrow 5r = 2n-7$. So $r = \frac{2n-7}{5}$. For the coefficient of $x^{11}$, we set $2n-5r' = 11 \Rightarrow 5r' = 2n-11$. So $r' = \frac{2n-11}{5}$. Since r and r' must be n

Question

Q.23. If the sum of the coefficients of  and  in the expansion of , , is zero, then the value of n is

FirstInTest · Accepted Answer

Sol. The general term in the expansion of $(x^2 + \frac{1}{x^3})^n$ is $T_{r+1} = \binom{n}{r} (x^2)^{n-r} (x^{-3})^r = \binom{n}{r} x^{2n-2r-3r} = \binom{n}{r} x^{2n-5r}$.
For the coefficient of $x^7$, we set $2n-5r = 7 \Rightarrow 5r = 2n-7$. So $r = \frac{2n-7}{5}$.
For the coefficient of $x^{11}$, we set $2n-5r' = 11 \Rightarrow 5r' = 2n-11$. So $r' = \frac{2n-11}{5}$.
Since r and r' must be non-negative integers, $2n-7$ and $2n-11$ must be multiples of 5.
Also, $0 \le r, r' \le n$.
The sum of the coefficients of $x^7$ and $x^{11}$ is zero.
$\binom{n}{r} + \binom{n}{r'} = 0$.
Since binomial coefficients $\binom{n}{r}$ are always positive for $n \ge r \ge 0$, their sum cannot be zero unless one of them is negative, which is not the case here. This implies that one of the coefficients must be negative, which happens if the term itself has a negative sign. However, the terms are $(x^2)$ and $(1/x^3)$, which are positive. So the coefficients $\binom{n}{r}$ are positive.
This condition implies that one of the coefficients is zero, or the problem implies that the terms themselves have alternating signs, which is not the case here. Let's assume the problem means the sum of the numerical values of the coefficients is zero, which means one coefficient must be negative. This can only happen if there's a negative sign in the original expression, which is not present.
Let's re-read the problem. "sum of the coefficients of $x^7$ and $x^{11}$ ... is zero". This usually implies that the coefficients are $\binom{n}{r}$ and $\binom{n}{r'}$ and their sum is zero. This is only possible if one of them is negative, which is not possible for $(a+b)^n$ where $a,b>0$. It could be $(a-b)^n$. But here it is $(x^2 + \frac{1}{x^3})^n$.
Perhaps the problem implies that $x^7$ and $x^{11}$ are not present in the expansion, meaning their coefficients are zero. If $\binom{n}{r} = 0$ and $\binom{n}{r'} = 0$, this is not possible for valid $r, r'$.
Let's consider the possibility that the problem implies that the coefficients are equal in magnitude but opposite in sign. This is not possible for this binomial expansion.
Let's consider the solution's approach. The solution simply states the answer is 21. This implies that the question is well-posed and has a unique integer solution for n. If the sum of coefficients is zero, it must be that one of the terms is not present or there is a misunderstanding of the question.
If the question meant $(x^2 - \frac{1}{x^3})^n$, then the general term would be $\binom{n}{r} (x^2)^{n-r} (-x^{-3})^r = \binom{n}{r} (-1)^r x^{2n-5r}$.
Then $\binom{n}{r_1}(-1)^{r_1} + \binom{n}{r_2}(-1)^{r_2} = 0$.
This would mean $\binom{n}{r_1}(-1)^{r_1} = -\binom{n}{r_2}(-1)^{r_2}$.
This implies $\binom{n}{r_1} = \binom{n}{r_2}$ and $(-1)^{r_1} = -(-1)^{r_2}$, which means $r_1$ and $r_2$ must have different parities.
Let's assume the problem is as stated and the coefficients are $\binom{n}{r}$ and $\binom{n}{r'}$. If their sum is zero, it implies that the coefficients are 0. This means $x^7$ and $x^{11}$ terms are not present in the expansion. This is not possible for a general $n$.
Let's assume the question implies that the sum of the coefficients is zero, which means that the terms $x^7$ and $x^{11}$ are not present in the expansion. This means that for the given $n$, there are no integer $r$ values that satisfy $2n-5r=7$ and $2n-5r=11$. This is also unlikely.
Let's consider the possibility that the problem is from a context where $n$ is a specific value. If $n=21$ (the answer), then:
For $x^7$: $5r = 2(21)-7 = 42-7 = 35 \Rightarrow r=7$. Coefficient is $\binom{21}{7}$.
For $x^{11}$: $5r' = 2(21)-11 = 42-11 = 31$. $r' = 31/5$, which is not an integer. So $x^{11}$ term is not present in the expansion if $n=21$. Its coefficient is 0.
If the coefficient of $x^{11}$ is 0, then the sum of coefficients of $x^7$ and $x^{11}$ is $\binom{21}{7} + 0 = \binom{21}{7}$. This is not zero.
There is a fundamental issue with the problem statement or my interpretation of "sum of the coefficients ... is zero" for this type of expansion. If the coefficients are always positive, their sum cannot be zero unless both are zero, which means the terms are not present.
Let's assume the problem implies that the sum of coefficients is zero, and the solution is 21. This implies that one of the coefficients is 0. If $n=21$, then $x^{11}$ is not present. So its coefficient is 0. But then the sum is $\binom{21}{7}$, which is not 0.
What if the problem meant $(x^2 - \frac{1}{x^3})^n$? Then $T_{r+1} = \binom{n}{r} (-1)^r x^{2n-5r}$.
For $x^7$: $r_1 = \frac{2n-7}{5}$. Coefficient $C_7 = \binom{n}{r_1}(-1)^{r_1}$.
For $x^{11}$: $r_2 = \frac{2n-11}{5}$. Coefficient $C_{11} = \binom{n}{r_2}(-1)^{r_2}$.
If $C_7 + C_{11} = 0$, then $\binom{n}{r_1}(-1)^{r_1} + \binom{n}{r_2}(-1)^{r_2} = 0$.
This requires $r_1$ and $r_2$ to have different parities and $\binom{n}{r_1} = \binom{n}{r_2}$.
If $\binom{n}{r_1} = \binom{n}{r_2}$, then either $r_1 = r_2$ (not possible as $7 
e 11$) or $r_1 + r_2 = n$.
So, $\frac{2n-7}{5} + \frac{2n-11}{5} = n$
$\frac{4n-18}{5} = n \Rightarrow 4n-18 = 5n \Rightarrow n = -18$. This is not possible as $n$ must be a positive integer.
Given the answer is 21, and the problem is from JEE Main, it's likely a standard type of question. The only way for the sum of positive coefficients to be zero is if both coefficients are zero. This means $x^7$ and $x^{11}$ terms are not present in the expansion.
If $n=21$, $r = (2(21)-7)/5 = 35/5 = 7$. So $x^7$ is present with coefficient $\binom{21}{7}$.
For $x^{11}$, $r' = (2(21)-11)/5 = 31/5$. Not an integer. So $x^{11}$ is not present, its coefficient is 0.
So the sum of coefficients is $\binom{21}{7} + 0 = \binom{21}{7}$. This is not zero.
There must be a typo in the question or the provided solution. However, I must extract the question as given and the answer as provided. I will set confidence to 0.7 and explanation to null due to this ambiguity.

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