The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:

Question

Aicharya · Accepted Answer

Solution: The general term T r+1 in the given expansion is given by ${^\text{45}}\text{C}_{\text{r}}\Big(4^{\frac{1}{5}}\Big)^{45-\text{r}}\Big(7^{\frac{1}{10}}\Big)^{\text{r}}$ For T r+1 to be an integer, we must have $\frac{\text{r}}{5}$ and $\frac{\text{r}}{10}$ as integers i.e. $0\leq\text{r}\leq45$ $\therefore \text{r}=0,10,20,30,40$ Hence, there are 5 rational and 41, i.e. 46 - 5, irrational terms. Alternative Solution: Sure, let's break it down. The general term of the binomial expansion $(a+b)^n$ is $\binom{n}{r} \cdot a^{n-r} \cdot b^r$ where $\binom{n}{r}$ is the binomial coefficient and $r$ is the term number. In the given expression, $a=4^{\frac{1}{5}}$ and $b=7^{\frac{1}{10}}$. For a term to be rational, the powers of $a$ and $b$ should both be integers. This means that $5 \cdot (n-r)$ and $10 \cdot r$ should both be integers, hence $n-r$ and $r$ should both be multiples of 5. The value of $r$ ranges from $0$ to $n=45$. The possible values of $r$ that are multiples of 5 are $0, 5, 10, 15, 20, 25, 30, 35, 40, 45$. So, there are 10 rational terms. The total number of terms in the expansion is 46 (from $r=0$ to $r=45$). Therefore, the number of irrational terms is $46-10 = 36$. However, this option is not given, so the correct answer is B) None of these.

The number of irrational terms in the expansion of \(\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}\) is:

Solution