Question

If k is and n be + ive integers and $s_k=1^k+2^k+3^k...+n^k,$ then show that
${\sum }_{r=1}^m{}^{m+1}{C}_r{S}_r=,(n+1{)}^{m+1}-(n+1).$Hence evaluate $s_4.$

Answer 1

$(1+x{)}^{m+1}={}^{m+1}{C}_0+{}^{m+1}{C}_{1}x+{}^{m+1}{C}_2{x}^2+.....{+}^{m+1}{C}_{m+1}{x}^{m+1}$
$therefore\left[\right(1+x{)}^{m+1}-x^{m+1}]-1={\sum }_{r=1}^m{}^{m+1}{C}_r{x}^r$
Now put x = 1,2,3,....,n and add and put
$1^r+2^r+3^r+....n^r=s_r$
L.H.S. is
$(2^{m+1}-1^{m+1})+(3^{m+1}-2^{m+1})+(4^{m+1}-3^{m+1})+..\left[\right(n+1{)}^{m+1}$
$-n^{m+1}]-\sum 1=(n+1{)}^{m+1}-1-\sum 1$
R.H.S. $={\sum }_{r=1}^m{}^{m+1}{C}_r{s}_r$
Put $\sum 1=n$
$\therefore (n+1{)}^{m+1}-(n+1)={\sum }_{r=1}^m$ ...(1)
Now putting m = 1 in (1),
$(n+1{)}^2-(n+1)={\sum }_{r=1}^1{}^2{C}_r{s}_r={}^2{C}_1\cdot s_1=2s_1$
$\therefore (n+1)\cdot ncdot\frac{1}{2}=s_1=1+2+3+..n$ .(2)
$(n+1{)}^3-(n+1)={\sum }_{r=1}^2{}^3{C}_r{s}_r={}^3{C}_1{s}_1+{}^3{C}_2{s}_2$
or $(n+1)\left[\right(n+1{)}^2-1]=3\cdot \frac{n(n+1)}{2}+3s_2$
or $(n+1)n\cdot (n+2)-\frac{3}{2}n(n+1)=3s_2$
or $n(n+1)(n+2-\frac{3}{2})=3s_2$
or $\frac{n(n+1)(2n+1)}{6}=s_2$ ..(3)
Now putting m = 3 in (1), we get
$(n+1{)}^4-(n+1)={\sum }_{r=1}^3{}^4{C}_r{s}_r$
$={}^4{C}_1{s}_1+{}^4{C}_2{s}_2+{}^4{C}_3{s}_3$
$=4\frac{n(n+1)}{2}+\frac{4.3}{1.2}\frac{n(n+1)(2n+1)}{6}+4s_3$
or $(n+1)\left[\right(n+1{)}^3-1-2n-n(2n+1)]=4s_3$
or $(n+1)(n^3+3n^2+3n+1-1-2n-2n^2-n)=4s_3$
or $(n+1)(n^3+n^2)=4s_3$
$\therefore s_3=\frac{1}{4}{n}^2(n+1{)}^2={\left[\frac{n(n+1)}{2}\right]}^2$
Lastly putting m = 4 in (1), we have
$(n+1{)}^5-(n+1)={\sum }_{r=1}^4{}^5{C}_r{s}_r$
$={}^5{C}_1{s}_1+{}^5{C}_2{s}_2+{}^5{C}_3{s}_3+{}^5{C}_4{s}_4$
$=5\cdot \frac{n(n+1)}{2}+10\cdot \frac{n(n+1)(2n+1)}{6}$ $+10\cdot \frac{n^2(n+1{)}^2}{4}+5\cdot s_4$
or $(n+1{)}^5-(n+1)-\frac{5}{2}n(n+1)$
$-\frac{5}{3}n(n+1)(2n+1)-\frac{5}{2}{n}^2(n+1{)}^2=5s_4$
$\frac{(n+1)}{6}\left[6\right(n+1{)}^4-6-15n+10(2n^2+n)$
$-15(n^3+n^2)]=5s_4$
$\therefore s_4=\frac{1}{30}(n+1)\left[6\right(n^4+4n^3+6n^2+4n+1)$
$-6-15n-20n^2-10n-15n^3-15n^2]$
$=\frac{1}{30}(n+1)[6n^4+9n^3+n^2-n]$
$=\frac{1}{30}n(n+1)(6n^3+9n^2+n-1)$
$=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$