Find the 4 th term of \( (x+5)^{7} \) \( \frac{625 x^{4}}{4375 x^{4}} \) \( \frac{21,875 x^{3}}{35 x^{4}} \)

To find the 4th term of \((x+5)^{7}\), we can use the binomial theorem, which states that the \( k \)-th term of \((a + b)^{n}\) is given by: \[ T_{k} = \binom{n}{k-1} a^{n-k+1} b^{k-1} \] In this case, \( a = x \), \( b = 5 \), and \( n = 7 \). For the 4th term, \( k = 4 \). Calculating \( T_{4} \): \[ T_{4} = \binom{7}{3} x^{7-4} (5)^{4-1} \] \[ = \binom{7}{3} x^{3} (5)^{3} \] Calculating \(\binom{7}{3}\): \[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] Now substituting this value back into \( T_{4} \): \[ = 35 x^{3} (125) = 35 \times 125 x^{3} = 4375 x^{3} \] Thus, the 4th term of \((x+5)^{7}\) is \( 4375 x^{3} \). As for the fraction calculations you provided: - \( \frac{625 x^{4}}{4375 x^{4}} = \frac{625}{4375} = \frac{1}{7} \) - \( \frac{21,875 x^{3}}{35 x^{4}} = \frac{21,875}{35}x^{-1} = 625 x^{-1} \), which simplifies the expression and is a great way to practice your fraction manipulation!