There are more then one representation of a series sum, like in case of "e" , there are many ways to represent "e" click here to know about them. We are providing you some of examples below.\(\bullet \displaystyle \sum _{k=0}^{n}k^2=1^{2} + 2^{2} + … + n^{2} = \displaystyle\frac{n(n+ 1)(2n+ 1)}{6}.\)
\(\bullet \displaystyle \sum _{k=0}^{n}k^3=1^{3} + 2^{3} + . . . . + n^{3}= \displaystyle \frac{{n^{2}(n + 1)^{2}}}{4 }\)
\(\bullet \displaystyle \sum _{k=1}^{n}(2k-1)=1 + 3 + ...... + (2n-1) =\displaystyle n^2 \)
\(\bullet \displaystyle \sum _{k=1}^{n}\frac{1}{k(k+1)}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots =\displaystyle\frac{n}{n+1} \)
\(\bullet \displaystyle \sum _{k=1}^{n} k(k+1)(k+2)=\displaystyle \frac{n(n+1)(n+2)(n+3){2}}{4} \)
\(\bullet \displaystyle \sum _{k=1}^{n}\frac{1}{k(k+1)(k+2)}=1 + 3 +\cdots+ n^{th}\;term =\displaystyle \frac{n(n+3)}{4(n+2)(n+3)} \)
\(\bullet \displaystyle \sum_{k=1}^{n}k^4=1+2^4+3^4+4^4+5^4+\cdots=\frac{6n^5+15n^4+10n^3-n}{30}\)
\(\bullet \displaystyle \sum _{n=0}^{\infty}=\frac{1}{2^{n}}=1+\frac{1}{2}+\frac{1}{4}+\cdots =2\)
\(\bullet \displaystyle \sum _{n=1}^{\infty }=\frac{\left(-1 \right)^{n-1} }{n}=1-\frac{1 }{ 2}+\frac{1 }{3}-\frac{1 }{4}\cdots =\ln(2)\quad\)
\(\bullet \displaystyle \sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n}}{2n-1}}= -1+{\frac {1}{3}}-{\frac {1}{5}}\cdots =-{\frac {\pi }{4}}\)
\(\bullet \displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}(4)}{2n-1}}={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}\cdots =\pi \)
\(\bullet \displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}\)
\(\bullet \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}=1+\frac {1}{1!}+\frac {1}{2!}+\frac {1}{3!}+\cdots =e \)
\(\bullet \displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}=1-{\frac {1}{1!}}+{\frac {1}{2!}}-{\frac {1}{3!}}+\cdots ={\frac {1}{e}}\)
0 Comments