Giải thích các bước giải:
1/. $a^3+b^3+c^3-3abc$
= $(a^3+b^3)+c^3-3abc$
= $(a+b)^3-3a^2b-3ab^2+c^3-3abc$
= $(a+b)^3-3ab(a+b)+c^3-3abc$
= $[(a+b)^3+c^3]-[3ab(a+b)-3abc]$
= $(a+b+c)[(a+b)^2-(a+b).c+c^2]-3ab(a+b+c)$
= [(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Mà $a+b+c=0$
⇒ $a^3+b^3+c^3-3abc=0$
⇒ $a^3+b^3+c3=3abc$ (đpcm)
2/. $2.2^3+3.2^4+4.2^5+5.2^6+...+n.2^{n+1}=2{n+5}$
Đặt A = $2.2^3+3.2^4+4.2^5+5.2^6+...+n.2^{n+1}$
2A = $2.2^4+3.2^5+4.2^6+5.2^7+...+(n-1).2^{n+1}+n.2^{n+2}$
2A-A = $2.2^4+3.2^5+4.2^6+5.2^7+...+(n-1).2^{n+1}+n.2^{n+2}-(2.2^3+3.2^4+4.2^5+5.2^6+...+n.2^{n+1})$
A = $-2^4-2^5-2^6-...-2^{n+1}+n.2^{n+2}-2.2^3$
A = $-(2^4+2^5+2^6+...+2^{n+1})+n.2^{n+2}-2^4$ (1)
Đặt E = $2^4+2^5+2^6+...+2^{n+1}$
2E = $2^5+2^6+2^7+...+2^{n+2}$
2E - E = $2^5+2^6+2^7+...+2^{n+2}-(2^4+2^5+2^6+...+2^{n+1})$
E = $2^n+2-2^4$ Thay vào (1)
A = $-(2^4+2^5+2^6+...+2^{n+1})+n.2^{n+2}-2^4$
A = $-(2^n+2-2^4)+n.2^{n+2}-2^4$
A = $2^4-2^n+2+n.2^{n+2}-2^4$
A = $n.2^{n+2}-2^n+2$
A = $.2^{n+2}(n-1)$
Mà: $.2^{n+2}(n-1)=2^{n+5}$
⇔ $.2^{n+2}(n-1)-2^{n+2}.2^3=0$
⇔ $.2^{n+2}(n-1-8)=0$
⇒ $.2^{n+2}(n-9)=0$
⇒ $2^{n+2}=0$ (không thỏa mãn) hay $n-9=0$ ⇒ $n=9$ (nhận)
⇒ $n=9$
Vậy $n=9$
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