1 Answer

Didier Prophete · Answer 1 · 2023-07-27T08:10:56+0000

Using the binomial, we have:

$(a+b)^n=\sum ^n_(k\mathop=0)(n!)/(k!(n-k)!)a^(n-k)b^k$

Here n = 5, then

$(a+b)^5=\sum ^5_{k\mathop{=}0}(5!)/(k!(5-k)!)a^(5-k)b^k$

The 4th term will be when k = 3, then

$(5!)/(3!(5-3)!)a^(5-3)b^3=(5!)/(3!\cdot2!)a^2b^3=10a^2b^3$

The 4th term will be

The correct answer is the letter C.

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