1. Find the sum of the products of every pair of the first n natural numbers
2. Sum the series : 1 + 4 + 10 + 22 + 46 + ...... to n terms.
3. Find the sum of first n terms of the series : 1(1)! + 2(2) ! + 3(3)! + 4(4)! + ........
213. The nth term = Tn = n(n) !
Tn can be written as
Tn = (n + 1 – 1) (n)!
Tn = (n + 1) ! – (n) ! ..........(i)
This is in the form f(n) – f(n – 1)
S = T1 + T2 + T3 + T4 + .......... + Tn
S = (2! – 1!) + (3! – 2!) + (4! – 3!) + ......... + {(n + 1)! – n!}
S = – 1! + (n + 1)!
→ S = (n + 1)! – 1
212.
The differences of successive terms are in G.P.
Let S = sum of first n terms.
⇒ S = 1 + 4 + 10 + 22 + 46 + .......... + Tn
⇒ S = 1 + 4 + 10 + 22 + 46 + ............ + Tn–1 + Tn
On subtracting, we get
0 = 1 + {3 + 6 + 12 + 24 + ............} – Tn
Tn = 1 + {3 + 6 + 12 + 24 + ........(n – 1) terms}
now you can solve
