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Answer 1 · 2017-11-26T12:18:44

Fifth term is $560 x^{3}$ $560x^3$

${(x + 2)}^{7}; t_{5} = ?$ $(x+2)^7 ; t_5 = ?$

We know $(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n$

Here $a=x,b=2,n=7$ We know, $nC_r = (n!)/(r!*(n-r)!$
$:.7C_4 =(7!)/(4!*3!)=35$

$t_5= nC_4*x^(n-4)*2^4$

$t_5= 35*x^(7-4)*2^4=560x^3$

Fifth term is $560x^3$ [Ans]

Answer 2 · 2017-11-26T12:55:07

$560x^3$

Refer to the file attached for an explanation.

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