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.
For a two-port network to be reciprocal, it is necessary that ……..