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
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The rank of matrix \(\begin{bmatrix} k & -1 & 0 \\[0.3em] 0 & k & -1 \\[0.3em] -1 & 0 & k \end{bmatrix}\) is 2, for \( k = \)
If \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\), then \(A^{-1} =\)
A two-port network is defined by the relation
\(\text{I}_1 = 5V_1 + 3V_2 \)
\(\text{I}_2 = 2V_1 - 7V_2 \)
The value of \( Z_{12} \) is:
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A 0 to 30 V voltmeter has an error of \(\pm 2\%\) of FSD. What is the range of readings if the voltage is 30V?