To find the fifth term in the binomial expansion of
(x + 1)58, we can use the binomial theorem. The binomial theorem states that:
(a + b)n =,
\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}
where \binom{n}{k} is a binomial coefficient, which can be calculated as:
\binom{n}{k} = \frac{n!}{k!(n-k)!}.
In our case, we have:
a = xb = 1n = 58
The k-th term in the expansion is given by:
T_{k+1} = \binom{n}{k} a^{n-k} b^{k}.
To find the fifth term, we need to set k = 4 (since counting starts from zero):
T_{5} = \binom{58}{4} x^{58-4} (1)^4
Now, we can calculate the binomial coefficient:
\binom{58}{4} = \frac{58!}{4!(58-4)!} = \frac{58!}{4! \, 54!} = \frac{58 \times 57 \times 56 \times 55}{4 \times 3 \times 2 \times 1}
Calculating this step by step:
- Calculate the numerator:
58 \times 57 \times 56 \times 55 = 9151920. - Calculate the denominator:
4 \times 3 \times 2 \times 1 = 24. - Now divide:
9151920 ÷ 24 = 381330.
So, \binom{58}{4} = 381330.
Next, we can substitute this back into our equation for the term:
T_{5} = 381330 \times x^{54} \times 1 = 381330 x^{54}.
Therefore, the fifth term in the binomial expansion of (x + 1)58 is: