Find the nth term of the series `1+2+5+12+25+46+"...."`.

Explanation

The sequence of first consecutive differences is `1,3,7,13,21,"…"`.The sequence of the second consecutive differences is `2,4,6,8,"..."`. Clearly, it is a AP. Then, nth term of the given series be
`T_(n)=a(n-1)(n-2)(n-3)+b(n-1)(n-2)+c(n-1)+d"......(i)"`
Putting `n=1,2,3,4` we get
`1=d " " "…..(ii)'`
`2=c+d" " "....(iii)"`
`5=2b+2c+d" " ".....(iv)"`
`12=6a+6b+3c+d" " ".....(v)"`
After, solving these equations, we get
`a=(1)/(3),b=1,c=1,d=1`
Putting the values of a,b,c,d in Eq. (i), we get
`T_(n)=(1)/(3)(n^(3)-6n^(2)+11n-6)+(n^(2)-3n+2)+(n-1)+1`
`=(1)/(3)(n^(3)-3n^(2)+5n)=(n)/(3)(n^(2)-3n+5)`