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Question: \[x^{5} + 10x^{4}a + 40x^{3}a^{2} + 80x^{2}a^{3} + 80xa^{4} + 32a^{5} =\]...

x5+10x4a+40x3a2+80x2a3+80xa4+32a5=x^{5} + 10x^{4}a + 40x^{3}a^{2} + 80x^{2}a^{3} + 80xa^{4} + 32a^{5} =

A

(x+a)5(x + a)^{5}

B

(3x+a)5(3x + a)^{5}

C

(−1)n/2(n+2)( - 1)^{n/2}(n + 2)

D

(x+2a)3(x + 2a)^{3}

Answer

(−1)n/2(n+2)( - 1)^{n/2}(n + 2)

Explanation

Solution

Conversely

(x+y)n=n⥂C0+n⥂C1xn−1y1+n⥂C2xn−2y2+....+nCnx0yn(x+2a)5(x + y)^{n} =^{n} ⥂ C_{0} +^{n} ⥂ C_{1}x^{n - 1}y^{1} +^{n} ⥂ C_{2}x^{n - 2}y^{2} + .... +^{n}C_{n}x^{0}y^{n}(x + 2a)^{5}

=5⥂C0x5+5⥂C1x4(2a)1+5⥂C2x3(2a)2+5C3x2(2a)3+5⥂C4x1(2a)4+5⥂C5x0(2a)55 ⥂ C_{0}x^{5} +^{5} ⥂ C_{1}x^{4}(2a)^{1} +^{5} ⥂ C_{2}x^{3}(2a)^{2} +^{5}C_{3}x^{2}(2a)^{3} +^{5} ⥂ C_{4}x^{1}(2a)^{4} +^{5} ⥂ C_{5}x^{0}(2a)^{5}

= x5+10x4a+40x3a2+80x2a3+80xa4+32a5x^{5} + 10x^{4}a + 40x^{3}a^{2} + 80x^{2}a^{3} + 80xa^{4} + 32a^{5}.