To determine how many terms will be in the expansion of \( (x + y + z)^5 \), we need to use the multinomial expansion formula. The number of terms in the expansion of \( (x + y + z)^n \) is given by the formula: \[ \binom{n+k-1}{k-1} \] where \( n \) is the exponent and \( k \) is the number of variables.
For the given expression \( (x + y + z)^5 \): - \( n = 5 \) (the exponent), - \( k = 3 \) (the number of variables: \( x \), \( y \), and \( z \)). The number of terms is therefore:
\[ \binom{5+3-1}{3-1} = \binom{7}{2} \] Now calculate \( \binom{7}{2} \): \[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]
Thus, there will be 21 terms in the expansion of \( (x + y + z)^5 \).
Conclusion:
The number of terms in the expansion of \( (x + y + z)^5 \) is 21.
The minimum value of $ n $ for which the number of integer terms in the binomial expansion $\left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n$ is 183, is
The sum of all rational terms in the expansion of \( \left( 1 + 2^{1/3} + 3^{1/2} \right)^6 \) is equal to
If \[ \sum_{r=0}^{10} \left( \frac{10^{r+1} - 1}{10^r} \right) \cdot {^{11}C_{r+1}} = \frac{\alpha^{11} - 11^{11}}{10^{10}}, \] then \( \alpha \) is equal to: