Question 1: Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Answer 1: The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n^th term, a_n = n (n + 1)

Question 2: Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5

+ …

Answer 2: The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n^th term,

a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

Question 3: Find the sum to n terms of the series 3 × 1² + 5 × 2² + 7 × 3² + …

Answer 3: The given series is 3 ×1² + 5 × 2² + 7 × 3² + … n^th term,

a_n = ( 2n + 1) n² = 2n³ + n²

Question 4: Find the sum to n terms of the series

Answer 4: The given series is

n^th term, a_n =    (By partial fractions)

Adding the above terms column wise, we obtain

Question 5: Find the sum to n terms of the series 5² + 6² + 7² + … + 20²

Answer 5: The given series is 5² + 6² + 7² + … + 20² n^th term,

a_n = ( n + 4)² = n² + 8n + 16

16^th term is (16 + 4)² = 20²

= 1496 + 1088 + 256

= 2840

∴ 5²+6²+7²+…+20² = 2840

Question 6: Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Answer 6: The given series is 3 × 8 + 6 × 11 + 9 × 14 + … a_n

= (n^th term of 3, 6, 9 …) × (n^th term of 8, 11, 14 …)

= (3n) (3n + 5)

= 9n² + 15n

= 3n(n+1)(n+3)

Question 7: Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …

Answer 7: The given series is 1² + (1² + 2²) + (1² + 2² + 3³ ) + … a_n

= (1² + 2² + 3³ +…….+ n²)

Question 8: Find the sum to n terms of the series whose n^th term is given by n (n + 1)(n + 4).

Answer 8: a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Question 9: Find the sum to n terms of the series whose nth terms is given by n² + 2ⁿ

Answer 9: an = n² + 2ⁿ

             (1)

Consider

The above series 2, 2², 2³ … is a G.P. with both the first term and common ratio equal to 2.

               (2)

Therefore, from (1) and (2), we obtain

Question 10: Find the sum to n terms of the series whose n^th terms is given by (2n – 1)²

Answer 10: a_n = (2n – 1)² = 4n² – 4n + 1