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 57955 by Sr@2004 last updated on 15/Apr/19

Commented by Sr@2004 last updated on 15/Apr/19

please solve 7,8

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

8)S_r =1+r+r^2 +r^3 +r^4 +...+r^(n−1) (r−1)S_r =(r−1)[1+r+r^2 +r^3 +...+r^(n−1) ] (r−1)S_r =(r−1)×((r^n −1)/(r−1)) (r−1)S_r =r^n −1 now value of[ S_2 +2S_3 +3S_4 +..+(n−1)S_n ] =Σ_(r=2) ^n (r−1)S_r =Σ_(r=2) ^n r^n −1 =(2^n −1)+(3^n −1)+(4^n −1)+...(n^n −1) =(2^n +3^n +..+n^n )+(−1)×(n−1) now value of S_1 =1+1+1+1...upto ntimes S_1 =1×n=n S_1 +(S_2 +2S_3 +3S_4 +...+(n−1)S_n ) =S_1 +(Σ_(r=2) ^n (r−1)S_r ) =S_1 +(2^n +3^n +4^n +...+n^n +(−1)(n−1) =n+(2^n +3^n +4^n +..+n^n )−n+1 =1+(2^n +3^n +4^n +...+n^n ) =1^n +2^n +3^n +...+n^n =proved

8)Sr=1+r+r2+r3+r4+...+rn−1(r−1)Sr=(r−1)[1+r+r2+r3+...+rn−1](r−1)Sr=(r−1)×rn−1r−1(r−1)Sr=rn−1nowvalueof[S2+2S3+3S4+..+(n−1)Sn]=∑nr=2(r−1)Sr=∑nr=2rn−1=(2n−1)+(3n−1)+(4n−1)+...(nn−1)=(2n+3n+..+nn)+(−1)×(n−1)nowvalueofS1=1+1+1+1...uptontimesS1=1×n=nS1+(S2+2S3+3S4+...+(n−1)Sn)=S1+(∑nr=2(r−1)Sr)=S1+(2n+3n+4n+...+nn+(−1)(n−1)=n+(2n+3n+4n+..+nn)−n+1=1+(2n+3n+4n+...+nn)=1n+2n+3n+...+nn=proved

Commented by Sr@2004 last updated on 16/Apr/19

and 7?

Terms of Service

Privacy Policy

Contact: info@tinkutara.com