Find the sum to n-terms of the sequence.
(i) 
+ 
+ 
+ ....
(ii)1 . 3 . 2 2 + 2 . 4 . 3 2 + 3 . 5 . 4 2 + ....
Text Solution
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(i) 
– 
(ii) 
(n + 1) (n + 2) (n + 3) (2n + 3)
Sol. (i) S = 
+ 
+ 
+ ......n terms; T n = 
∴ T n = 
; T 1 = 
, T 2 = 
,
T 3 = 
T n = 
sum of all terms gives S n
⇒ S n = 
(ii) 1.3.2 2 + 2.4.3 2 + 3.5.4 2 + ......... n terms
T n = n(n + 2) (n + 1) 2 = n(n + 1) (n + 2) (n + 3 – 2)
T n = n (n+1) (n+2) (n+3) – 2(n) (n+1) (n + 2)
S n = S 1 – 2S 2
S 1 = 
= 
– (r – 1) r (r + 1) (r + 2) (r + 3)
= 
[1.2.3.4.5 – 0] + 
[2.3.4.5.6 – 1.2.3.4.5] + 
[3.4.5.6.7 – 2.3.4.5.6] +..
.......... + 
[n(n +1) (n + 2) (n + 3) (n + 4) – (n – 1) n (n + 1) (n + 2) (n + 3)]
∴ S 1 = 
[n (n + 1) (n + 2) (n + 3) (n + 4)]
Now S 2 = 
= 
= 
[1.2.3.4 – 0] + 
[2.3.4.5 – 1.2.3.4] + 
[3.4.5.6 – 2.3.4.5]
......... + 
[n(n + 1) (n + 2) (n + 3) – (n – 1) n (n + 1) (n + 2)]
∴ S 2 = 
[n (n + 1) (n + 2) (n + 3)]
∴ S n = 
– 
[n(n + 1) (n + 2) (n + 3)]
= n (n + 1) (n + 2) (n + 3) 
= 
n (n + 1) (n + 2) (n + 3) (2n + 3)
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