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](Q57972.png)
$$\left.\mathrm{8}\right){S}_{{r}} =\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +{r}^{\mathrm{4}} +...+{r}^{{n}−\mathrm{1}} \\ $$$$\left({r}−\mathrm{1}\right){S}_{{r}} =\left({r}−\mathrm{1}\right)\left[\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +...+{r}^{{n}−\mathrm{1}} \right] \\ $$$$\left({r}−\mathrm{1}\right){S}_{{r}} =\left({r}−\mathrm{1}\right)×\frac{{r}^{{n}} −\mathrm{1}}{{r}−\mathrm{1}} \\ $$$$\left({r}−\mathrm{1}\right){S}_{{r}} ={r}^{{n}} −\mathrm{1} \\ $$$${now}\:{value}\:{of}\left[\:{S}_{\mathrm{2}} +\mathrm{2}{S}_{\mathrm{3}} +\mathrm{3}{S}_{\mathrm{4}} +..+\left({n}−\mathrm{1}\right){S}_{{n}} \right] \\ $$$$=\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\left({r}−\mathrm{1}\right){S}_{{r}} \\ $$$$=\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}{r}^{{n}} −\mathrm{1} \\ $$$$=\left(\mathrm{2}^{{n}} −\mathrm{1}\right)+\left(\mathrm{3}^{{n}} −\mathrm{1}\right)+\left(\mathrm{4}^{{n}} −\mathrm{1}\right)+...\left({n}^{{n}} −\mathrm{1}\right) \\ $$$$=\left(\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +..+{n}^{{n}} \right)+\left(−\mathrm{1}\right)×\left({n}−\mathrm{1}\right) \\ $$$${now}\:{value}\:{of}\:{S}_{\mathrm{1}} =\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}...{upto}\:{ntimes} \\ $$$${S}_{\mathrm{1}} =\mathrm{1}×{n}={n} \\ $$$${S}_{\mathrm{1}} +\left({S}_{\mathrm{2}} +\mathrm{2}{S}_{\mathrm{3}} +\mathrm{3}{S}_{\mathrm{4}} +...+\left({n}−\mathrm{1}\right){S}_{{n}} \right) \\ $$$$={S}_{\mathrm{1}} +\left(\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\left({r}−\mathrm{1}\right){S}_{{r}} \right) \\ $$$$={S}_{\mathrm{1}} +\left(\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +...+{n}^{{n}} +\left(−\mathrm{1}\right)\left({n}−\mathrm{1}\right)\right. \\ $$$$={n}+\left(\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +..+{n}^{{n}} \right)−{n}+\mathrm{1} \\ $$$$=\mathrm{1}+\left(\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +...+{n}^{{n}} \right) \\ $$$$=\mathrm{1}^{{n}} +\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +...+{n}^{{n}} \\ $$$$={proved} \\ $$$$ \\ $$
Commented by Sr@2004 last updated on 16/Apr/19
and 7?