Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 126617 by Khalmohmmad last updated on 22/Dec/20

1∙2+2∙3+3∙4+...+n(n+1) S_n =?

1⋅2+2⋅3+3⋅4+...+n(n+1)Sn=?

Answered by Dwaipayan Shikari last updated on 22/Dec/20

Σ_(n=1) ^n n^2 +n = ((n(n+1)(2n+1))/6)+((n(n+1))/2)=((n(n+1))/2).((2n+4)/3) =((n(n+1)(n+2))/3)

∑nn=1n2+n=n(n+1)(2n+1)6+n(n+1)2=n(n+1)2.2n+43=n(n+1)(n+2)3

Commented by JDamian last updated on 22/Dec/20

Review the sum index

Reviewthesumindex

Answered by Bird last updated on 23/Dec/20

S_n =Σ_(k=1) ^n k(k+1) =Σ_(k=1) ^n k^(2 ) +Σ_(k=1) ^n k =((n(n+1)(2n+1))/6) +((n(n+1))/2) =((n(n+1))/2)(((2n+1)/3)+1) =((n(n+1))/2)(((2n+4)/3))=((n(n+1)(n+2))/3)

Sn=∑k=1nk(k+1)=∑k=1nk2+∑k=1nk=n(n+1)(2n+1)6+n(n+1)2=n(n+1)2(2n+13+1)=n(n+1)2(2n+43)=n(n+1)(n+2)3

Terms of Service

Privacy Policy

Contact: info@tinkutara.com