Lời giải:
Áp dụng hằng đẳng thức đáng nhớ:
\((1+1)^4=1^4+4.1^3+6.1^2+4.1+1\)
\((2+1)^4=2^4+4.2^3+6.2^2+4.2+1\)
\((3+1)^4=3^4+4.3^3+6.3^2+4.3+1\)
..............
\((n+1)^4=n^4+4.n^3+6n^2+4.n+1\)
Cộng theo vế:
\(2^4+3^4+...+(n+1)^4=(1^4+2^4+...+n^4)+4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+n\)
\((n+1)^4=4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+n+1(1)\)
----------------------------
\((1+1)^3=1^3+3.1^2+3.1+1\)
\((2+1)^3=2^3+3.2^2+3.2+1\)
......
\((n+1)^3=n^3+3.n^2+3.n+1\)
Cộng theo vế:
\(2^3+3^3+...+(n+1)^3=(1^3+2^3+...+n^3)+3(1^2+2^2+...+n^2)+3(1+2+...+n)+n\)
\((n+1)^3=3(1^2+2^2+...+n^2)+3(1+2+...+n)+n+1\)
\(1^2+2^2+...+n^2=\frac{(n+1)^3-3(1+2+..+n)-n-1}{3}(2)\)
Từ \((1);(2)\Rightarrow (n+1)^4=4(1^3+2^3+...+n^3)+2[(n+1)^3-3(1+2+..+n)-n-1]+4(1+2+...+n)+n+1\)
\(=4(1^3+2^3+..+n^3)+2(n+1)^3-2(1+2+...+n)-(n+1)\)
\(\Rightarrow 1^3+2^3+...+n^3=\frac{(n+1)^4-2(n+1)^3+2(1+2+..+n)+n+1}{4}=\frac{n^4+2n^3+n^2}{4}=\frac{n^2(n+1)^2}{4}\)
\(\Rightarrow \sqrt{1^3+2^3+...+n^3}=\frac{n(n+1)}{2}=1+2+...+n\) (đpcm)
Lời giải:
Áp dụng hằng đẳng thức đáng nhớ:
\((1+1)^4=1^4+4.1^3+6.1^2+4.1+1\)
\((2+1)^4=2^4+4.2^3+6.2^2+4.2+1\)
\((3+1)^4=3^4+4.3^3+6.3^2+4.3+1\)
..............
\((n+1)^4=n^4+4.n^3+6n^2+4.n+1\)
Cộng theo vế:
\(2^4+3^4+...+(n+1)^4=(1^4+2^4+...+n^4)+4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+n\)
\((n+1)^4=4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+n+1(1)\)
----------------------------
\((1+1)^3=1^3+3.1^2+3.1+1\)
\((2+1)^3=2^3+3.2^2+3.2+1\)
......
\((n+1)^3=n^3+3.n^2+3.n+1\)
Cộng theo vế:
\(2^3+3^3+...+(n+1)^3=(1^3+2^3+...+n^3)+3(1^2+2^2+...+n^2)+3(1+2+...+n)+n\)
\((n+1)^3=3(1^2+2^2+...+n^2)+3(1+2+...+n)+n+1\)
\(1^2+2^2+...+n^2=\frac{(n+1)^3-3(1+2+..+n)-n-1}{3}(2)\)
Từ \((1);(2)\Rightarrow (n+1)^4=4(1^3+2^3+...+n^3)+2[(n+1)^3-3(1+2+..+n)-n-1]+4(1+2+...+n)+n+1\)
\(=4(1^3+2^3+..+n^3)+2(n+1)^3-2(1+2+...+n)-(n+1)\)
\(\Rightarrow 1^3+2^3+...+n^3=\frac{(n+1)^4-2(n+1)^3+2(1+2+..+n)+n+1}{4}=\frac{n^4+2n^3+n^2}{4}=\frac{n^2(n+1)^2}{4}\)
\(\Rightarrow \sqrt{1^3+2^3+...+n^3}=\frac{n(n+1)}{2}=1+2+...+n\) (đpcm)
\(\Leftrightarrow1^3+2^3+........+n^3=\left(1+2+3+......n\right)^2\)
\(\left(1+2+3+....+n\right)^2-\left[1+2+3+.....+\left(n-1\right)\right]^2=\frac{\left(n+1\right)^2n^2}{4}-\frac{n^2\left(n-1\right)^2}{4}=\frac{n^2.4n}{4}=n^3\)\(tuongtu:\left[1+2+3+....+\left(n-1\right)\right]^2-\left[1+2+3+.....+\left(n-2\right)\right]^2=\left(n-1\right)^3\) \(....\)
\(1^2-0^2=1^3\)
\(Congvetheove\Rightarrow\left(1+2+....+n\right)^2-\left[1+2+...+\left(n-1\right)\right]^2+\left[1+2+...+\left(n-1\right)\right]^2-\left[1+2+.....+\left(n-2\right)\right]^2+.....-\left(0+1\right)^2+\left(0+1\right)^2-0^2=1^3+2^3+.....+n^3\Rightarrow\left(1+2+3+.....+n\right)^2=1^3+2^3+.....+n^3\left(dpcm\right)\)
