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Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
nth term, an = n ( n + 1)
Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
nth term, an = n ( n + 1) ( n + 2)
= (n2 + n) (n + 2)
= n3 + 3n2 + 2n
Find the sum to n terms of each of the series 3 × 12 + 5 × 22 + 7 × 32 + …
The given series is 3 ×12 + 5 × 22 + 7 × 32 + …
nth term, an = ( 2n + 1) n2 = 2n3 + n2
Find the sum to n terms of the series 
The given series is
nth term, an = 
Adding the above terms column wise, we obtain
Find the sum to n terms of the series 
The given series is 52 + 62 + 72 + … + 202
nth term, an = ( n + 4)2 = n2 + 8n + 16
16th term is (16 + 4)2 = 2022
Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…
The given series is 3 × 8 + 6 × 11 + 9 × 14 + …
an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)
= (3n) (3n + 5)
= 9n2 + 15n
Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
The given series is 12 + (12 + 22) + (12 + 22 + 32 ) + …
an = (12 + 22 + 32 +…….+ n2)
Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n
Find the sum to n terms of the series whose nth terms is given by n2 + 2n
an = n2 + 2n
Consider 
The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.
Therefore, from (1) and (2), we obtain
Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
an= (2n – 1)2 = 4n2 – 4n + 1
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