Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 147878 by Tawa11 last updated on 24/Jul/21

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{to}\mathrm{n}\mathrm{term}:1.2.3+4.5.6+7.8.9+...$$

Answered by Olaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{S}=\underset{k=1}{\sum ^n}(3k-2)(3k-1)\left(3k\right)$$$$\mathrm{S}=3\underset{k=1}{\sum ^n}(9k^3-9k^2+2k)$$$$\mathrm{S}=3(9\frac{n^2{(n+1)}^2}{4}-9\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2})$$$$\mathrm{S}=3n(n+1)(9\frac{n(n+1)}{4}-9\frac{(2n+1)}{6}+1)$$$$\mathrm{S}=\frac{3}{4}n(n+1)(9n^2-3n-2)$$$$\mathrm{S}=\frac{3}{4}n(n+1)(3n-2)(3n+1)$$

Commented by puissant last updated on 24/Jul/21

$$\mathrm{monsieur}\mathrm{au}\mathrm{debut}{\mathrm{c}}^{\prime }\mathrm{est}\mathrm{k}\mathrm{qui}\mathrm{varie}$$$$\mathrm{mais}\mathrm{vous}\mathrm{avez}\mathrm{mis}\mathrm{n}..\mathrm{on}\mathrm{va}\mathrm{supposer}$$$$\mathrm{que}{\mathrm{c}}^{\prime }\mathrm{est}\mathrm{une}\mathrm{erreur}..$$

Commented by Tawa11 last updated on 24/Jul/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciste}.$$

Commented by Olaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{bien}\mathrm{vu}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com