Формулы для нахождения сумм
- \(1+2+3+…+n=\displaystyle\frac{n(n+1)}{2}\)
- \(p+(p+1)+(p+2)+…+(p+n)=\displaystyle\frac{(n+1)(2p+n)}{n}\)
- \(1^2+2^2+3^2+…+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}\)
- \(1^3+2^3+3^3+…+n^3=\displaystyle\frac{n^2(n+1)^2}{4}\)
- \(1^4+2^4+3^4+…+n^4=\displaystyle\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\)
- \(1+3+5+…+(2n-1)=n^2\)
- \(1^2+3^2+5^2+…+(2n-1)^2=\displaystyle\frac{n(4n^2-1)}{3}\)
- \(1^3+3^3+5^3+…+(2n-1)^3=n^2(2n^2-1)\)
- \(1\cdot 2+2\cdot 3+…+n(n+1)=\displaystyle\frac{n(n+1)(n+2)}{3}\)
- \(1\cdot 4+2\cdot 7+…+n(3n+1)=n(n+1)^2\)
- \(1\cdot 2\cdot 3+2\cdot 3\cdot 4+…+n(n+1)(n+2)=\displaystyle\frac{1}{4}n(n+1)(n+2)(n+3)\)
- \(1^2-2^2+3^2-4^2+…+(-1)^{n-1}n^2=(-1)^{n-1}\displaystyle\frac{n(n+1)}{2}\)
- \(\displaystyle\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+…+\frac{1}{n(n+1)}=\displaystyle\frac{n}{n+1}\)
- \(\displaystyle\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+…+\frac{1}{(2n-1)(2n+1)}=\displaystyle\frac{n}{2n+1}\)
- \(\sin x+\sin 2x+\sin 3x+…+\sin nx=\displaystyle\frac{\sin\frac{n+1}{2}x\cdot\sin\frac{nx}{2}}{\sin\frac{x}{2}}\)
- \(\cos x+\cos 2x+\cos 3x+…+\cos nx=\displaystyle\frac{\cos\frac{n+1}{2}x\cdot\sin\frac{nx}{2}}{\sin\frac{x}{2}}\)