Respuesta :

konrad509

[tex]\displaystyle\\(a+b)^n\\T_{r+1}=\binom{n}{r}a^{n-r}b^r\\\\\\(x+2)^7\\a=x\\b=2\\r+1=5\Rightarrow r=4\\n=7\\T_5=\binom{7}{4}x^{7-4}2^4\\T_5=\dfrac{7!}{4!3!}\cdot x^3\cdot16\\T_5=16\cdot \dfrac{5\cdot6\cdot7}{2\cdot3}\cdot x^3\\\\T_5=560x^3[/tex]

09pqr4sT

Answer:

[tex]\large \boxed{560x^3}[/tex]

Step-by-step explanation:

[tex](x+2)^7[/tex]

Expand brackets.

[tex](x+2) (x+2) (x+2) (x+2) (x+2) (x+2) (x+2)[/tex]

[tex](x^2 +4x+4) (x^2 +4x+4) (x^2 +4x+4)(x+2)[/tex]

[tex](x^4 +8x^3 +24x^2 +32x+16)(x^3 +6x^2 +12x+8)[/tex]

[tex]x^7 +14x^6 +84x^5 +280x^4 +560x^3 +672x^2 +448x+128[/tex]

The fifth term is 560x³.

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