Question 1198571

Given that 

{{{f(x)=(r+1)!/(r-1)!}}}

Simplify:

{{{f(x)=(r+1)!/(r-1)!}}}

cancel the factorials: {{{n!/(n-m)!=n(n-1)}}}*......*{{{(n-m+1)}}}, {{{n>m}}}

{{{(r+1)!/(r-1)!=r(r+1)}}}

siplified:

{{{f(x)=r (r + 1)}}}


Evaluate the sum from {{{1}}} to {{{10}}} of {{{f(x)}}}

{{{f(1)=1 (1+ 1)=2}}}

{{{f(2)=2 (2+ 1)=6}}}

{{{f(3)=3 (3+ 1)=12}}}
{{{f(4)=4 (4+ 1)=20}}}

{{{f(5)=5 (5+ 1)=30}}}
{{{f(6)=6 (6+ 1)=42}}}
{{{f(7)=7 (7+ 1)=56}}}
{{{f(8)=8 (8+ 1)=72}}}
{{{f(9)=9 (9+ 1)=90}}}
{{{f(10)=10 (10+ 1)=110}}}

the sum is;

{{{2+6+12+20+30+42+56+72+90+110=440}}}