Observe the given binomial expression and exponent.
Step 2 :
Apply the formal expression of the binomial theorem:
(a + b)n = $\sum_{k = 0}^{n}$ $\frac{n!}{(n - k)! k!}$ an - k bk
Expand: (x + 2)6
Step 1 :Given that: a = x, b = 2 and n = 6
Step 2 :(x + 2)6 = x6 + $\frac{6 \times 2 \times x^{5}}{1!}$ + $\frac{6 \times 5 \times x^{4} \times 2^{2}}{2!}$ + $\frac{6 \times 5 \times 4 \times x^{3} \times 2^{3}}{3!}$ + $\frac{6 \times 5 \times 4 \times 3 \times x^{2} \times 2^{4}}{4!}$ + $\frac{6 \times 5 \times 4 \times 3 \times 2 \times x^{1} \times 2^{5}}{5!}$ + 26
=> x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64
Answer :(x + 2)6 = x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64
Expand: (y + 4)6
Step 1 :Given that: a = y, b = 4 and n = 6
Step 2 :(y + 4)6 = y6 + $\frac{6 \times 4 \times y^{5}}{1!}$ + $\frac{6 \times 5 \times y^{4} \times 4^{2}}{2!}$ + $\frac{6 \times 5 \times 4 \times y^{3} \times 4^{3}}{3!}$ + $\frac{6 \times 5 \times 4 \times 3 \times y^{2} \times 4^{4}}{4!}$ + $\frac{6 \times 5 \times 4 \times 3 \times 2 \times y^{1} \times 4^{5}}{5!}$ + 46
=> y6 + 24y5 + 240y4 + 1280y3 + 3840y2 + 6144y + 4096
Answer :(y + 4)6 = y6 + 24y5 + 240y4 + 1280y3 + 3840y2 + 6144y + 4096
