Câu hỏi của Nguyễn Mai Oanh

Question

tính nhanh :1 mũ 3+2 mũ 3+3 mũ 3 +4 mũ 3+5 mũ 3

Bạch Tố Như · Accepted Answer

=15$^3$Câu trả lời: =15$^3$

Lê Tài Bảo Châu · Answer

Mình làm bài tổng quát nha để bạn hiểu sau rồi bạn thay vào . Đặt $S_1=1+2+...+n$$\Rightarrow S_1=\frac{n\left(n+1\right)}{2}$Đặt $S_2=1^2+2^2+...+n^2$Ta có: $2^3=\left(1+1\right)^3=1^3+3.1^2+3.1+1$$3^3=\left(2+1\right)^3=2^3+3.2^2+3.2+1$..................................................................................$\left(n+1\right)^3=n^3+3n^2+3n+1$Cộng từng vế n thẳng đẳng thức trên ta được :$\left(n+1\right)^3=1^3+3.\left(1^2+2^2+...+n^2\right)+3.\left(1+2+3+...+n\right)+n$$\Rightarrow\left(n+1\right)^3=1^3+3.\left(1^2+2^2+...+n^2\right)+\frac{3n\left(n+1\right)}{2}+n$$\Rightarrow3.\left(1^2+2^2+...+n^2\right)=\left(n+1\right)^3-\frac{3n\left(n+1\right)}{2}-\left(n+1\right)$Hay $3S_2=\left(n+1\right)\left[\left(n+1\right)^2-\frac{3n}{2}-1\right]$$\Rightarrow3S_2=\left(n+1\right)\left(n^2+\frac{n}{2}\right)$$\Rightarrow3S_2=\frac{1}{2}n\left(n+1\right)\left(2n+1\right)$$\Rightarrow S_2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$Đặt $S_3=1^3+2^3+...+n^3$Ta có: $\left(1+1\right)^4=1^4+4.1^3+6.1^2+4.1+1$$\left(2+1\right)^4=2^4+4.2^3+6.2^2+4.2+1$........................................................................................$\left(n+1\right)^4=n^4+4n^3+6n^2+4n+1$Cộng từng vế n đẳng thức trên ta được :$\left(n+1\right)^4=1^4+4.\left(1^3+2^3+...+n^3\right)+6.\left(1^2+2^2+...+n^2\right)+4.\left(1+2+...+n\right)+n$$\Rightarrow\left(n+1\right)^4=1+4S_3+6S_2+4S_1+n$Đã chứng minh $S_1=\frac{n\left(n+1\right)}{2}$$S_2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$Từ đó tính được :$S_3=\frac{n^2\left(n+1\right)^2}{4}$đó là công thức giờ chỉ vệc thay vào$1^3+2^3+3^3+4^3+5^3=\frac{5^2\left(5+1\right)^2}{4}=225$Câu trả lời: Mình làm bài tổng quát nha để bạn hiểu sau rồi bạn thay vào . Đặt $S_1=1+2+...+n$$\Rightarrow S_1=\frac{n\left(n+1\right)}{2}$Đặt $S_2=1^2+2^2+...+n^2$Ta có: $2^3=\left(1+1\right)^3=1^3+3.1^2+3.1+1$$3^3=\left(2+1\right)^3=2^3+3.2^2+3.2+1$..................................................................................$\left(n+1\right)^3=n^3+3n^2+3n+1$Cộng từng vế n thẳng đẳng thức trên ta được :$\left(n+1\right)^3=1^3+3.\left(1^2+2^2+...+n^2\right)+3.\left(1+2+3+...+n\right)+n$$\Rightarrow\left(n+1\right)^3=1^3+3.\left(1^2+2^2+...+n^2\right)+\frac{3n\left(n+1\right)}{2}+n$$\Rightarrow3.\left(1^2+2^2+...+n^2\right)=\left(n+1\right)^3-\frac{3n\left(n+1\right)}{2}-\left(n+1\right)$Hay $3S_2=\left(n+1\right)\left[\left(n+1\right)^2-\frac{3n}{2}-1\right]$$\Rightarrow3S_2=\left(n+1\right)\left(n^2+\frac{n}{2}\right)$$\Rightarrow3S_2=\frac{1}{2}n\left(n+1\right)\left(2n+1\right)$$\Rightarrow S_2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$Đặt $S_3=1^3+2^3+...+n^3$Ta có: $\left(1+1\right)^4=1^4+4.1^3+6.1^2+4.1+1$$\left(2+1\right)^4=2^4+4.2^3+6.2^2+4.2+1$........................................................................................$\left(n+1\right)^4=n^4+4n^3+6n^2+4n+1$Cộng từng vế n đẳng thức trên ta được :$\left(n+1\right)^4=1^4+4.\left(1^3+2^3+...+n^3\right)+6.\left(1^2+2^2+...+n^2\right)+4.\left(1+2+...+n\right)+n$$\Rightarrow\left(n+1\right)^4=1+4S_3+6S_2+4S_1+n$Đã chứng minh $S_1=\frac{n\left(n+1\right)}{2}$$S_2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$Từ đó tính được :$S_3=\frac{n^2\left(n+1\right)^2}{4}$đó là công thức giờ chỉ vệc thay vào$1^3+2^3+3^3+4^3+5^3=\frac{5^2\left(5+1\right)^2}{4}=225$

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