Find the nth term of the series `1+5+18+58+179+"..."`.

Explanation

The sequence of first consctive differences is `4,13,40,121,"…"` and second consecutive differences is `9,27,81,"….",`. Clearly, it is an GP with comon ratio 3. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then,
`S_(n)=1+5+18+58+179+".."+T_(n-1)+T_(n)"....(i)"`
`S_(n)=1+5+18+58+".."+T_(n-1)+T_(n)"....(ii)"`
Subtracting Eq. (ii) from Eq.(i), we get
`0=1+4+13+40+121+".."+(T_(n)-T_(n))-T_(n)`
`implies T_(n)=1+4+13+40+121+".." " upto n terms "`
or ` T_(n)=1+4+13+40+121+".."+t_(n-1)+t_(n)"....(iii)"`
` T_(n)=1+4+13+40+".."+t_(n-1)+t_(n)"....(iv)"`
Now, subtracting Eq. (iv) from Eq. (iii), we get ` 0=1+3+9+27+81+".."+(t_(n)-t_(n-1))-t_(n)`
or `t_(n)=1+3+9+27+81+".." " upto n terms "`
`(1*(3^(n)-1))/((3-1))=(1)/(2)(3^(n)-1)`
` therefore T_(n)=sumt_(n)=(1)/(2)(sum3^(n)-sum1)`
`=(1)/(2){(3+3^(2)+3^(3)+"..."+3^(n))-n}`
`=(1)/(2){(3(3^(n)-1))/((3-1))-n}`
`=(3)/(4)(3^(n)-1)-(1)/(2)n`