Answer:
[tex]{x}^{5} + 40{x}^{4} + 640{x}^{3} + 5120 {x}^{2} + 20480{x}^{}+ 32768 [/tex]
Step-by-step explanation:
This too follows the as same process as
recall binomial theorem,
[tex] \displaystyle (a + b {)}^{n} = \sum _{k = 0} ^{n} \binom{n}{k} {a}^{n - k} {b}^{k} [/tex]
Given binomial:
comparing it to (a+b)ⁿ , we acquire:
Now substitute:
[tex] \displaystyle (x +8 {)}^{5} = \sum _{k = 0} ^{5} \binom{n}{k} {a}^{5- k} {b}^{k} [/tex]
To convert the summation into a sum, substitute the values from k=0 to k=5 into the expression which yields
[tex] \displaystyle (x + 8{)}^{5} = \binom{5}{0} {x}^{5 - 0} \cdot {8}^{0} + \binom{5}{1} {x}^{5 - 1} \cdot {8}^{1} + \binom{5}{2} {x}^{5 - 2} \cdot {8}^{2} + \binom{5}{3} {x}^{5 - 3} \cdot {8}^{3} + \binom{5}{4} {x}^{5- 4} \cdot {8}^{4} + \binom{5}{5} {x}^{5 - 5} \cdot {8}^{5} \\ \implies(x + 8 {)}^{5} = \binom{5}{0} {x}^{5} \cdot 1 + \binom{5}{1} {x}^{4} \cdot {8} + \binom{5}{2} {x}^{3} \cdot 64+ \binom{5}{3} {x}^{2} \cdot 512 + \binom{6}{4} {x}^{1} \cdot 4096+ \binom{5}{5} {x}^{0} \cdot 32768\\ \implies(x + 8{)}^{5} = 1 \cdot {x}^{5 } \cdot 1 + 5\cdot{x}^{4} \cdot {8} + 10 \cdot {x}^{3} \cdot 64+ 10 \cdot {x}^{2} \cdot 512 + 5 \cdot{x}^{1} \cdot 4096+ 1 \cdot {x}^{0} \cdot 32768 \\ \implies \boxed{ (x + 8 {)}^{5} = {x}^{5} + 40{x}^{4} + 640{x}^{3} + 5120 {x}^{2} + 20480{x}^{}+ 32768 }[/tex]
and we're done!
