### Find the sum to n terms of each of the series in Exercises 1 to 7.

### Question 1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

**Solution:**

Given:Series = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …To find nth term, we have

n

^{th}term, a_{n}= n ( n + 1)So, the sum of n terms of the series:

### Question 2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

**Solution:**

Given:Series = 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …To find nth term, we have

n

^{th}term, a_{n}= n(n + 1)(n + 2)= (n

^{2}+ n) (n + 2)= n

^{3}+ 3n^{2}+ 2nSo, the sum of n terms of the series:

### Question 3. 3 × 1^{2} + 5 × 2^{2} + 7 × 3^{2} + …

**Solution:**

Given:Series = 3 × 1^{2}+ 5 × 2^{2}+ 7 × 3^{2}+ …To find nth term, we have

n

^{th}term, a_{n}= (2n + 1) n^{2}= 2n^{3}+ n^{2}So, the sum of n terms of the series:

### Question 4. Find the sum to n terms of the series

**Solution:**

Given:Series =To find nth term, we have

n

^{th}term a_{n}= (By partial fractions)So, on adding the above terms columns wise, we obtain

### Question 5. Find the sum to n terms of the series 5^{2} + 6^{2} + 7^{2} + … + 20^{2}

**Solution:**

Given:Series = 5^{2}+ 6^{2}+ 7^{2}+ … + 20^{2}To find nth term, we have

n

^{th}term, a_{n}= (n + 4)^{2}= n^{2}+ 8n + 16So, the sum of n terms of the series:

16

^{th}term is (16 + 4) = 20^{2}= 1496 + 1088 + 256

= 2840

So, 5

^{2}+ 6^{2}+ 7^{2}+ …..+ 20^{2}= 2840

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

**Solution:**

Given:Series = 3 × 8 + 6 × 11 + 9 × 14 + …To find nth term, we have

a

_{n}= (n^{th}term of 3, 6, 9 …) × (n^{th}term of 8, 11, 14, …)= (3n) (3n + 5)

= 9n

^{2}+ 15nSo, the sum of n terms of the series:

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

### Question 7. Find the sum to n terms of the series 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + …

**Solution:**

Given:Series = 1^{2}+ (1^{2}+ 2^{2}) + (1^{2}+ 2^{2}+ 3^{2)}+ …To find nth term, we have

a

_{n}= (1^{2}+ 2^{2}+ 3^{2}+…….+ n^{2})So, the sum of n terms of the series:

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

**Solution:**

Given:a_{n}= n (n + 1) (n + 4) = n(n^{2}+ 5n + 4) = n^{3}+ 5n^{2}+ 4nNow, the sum of n terms of the series:

### Question 9. Find the sum to n terms of the series whose nth terms is given by n^{2} + 2^{n}

**Solution:**

Given:n^{th}term of the series as:a

_{n}= n^{2}+ 2^{n}So, the sum of n terms of the series:

Now, the above series 2, 2

^{2}, 2^{3}….. is G.P.and the first term and common ration is equal to 2.

From eq(1) and (2), we get

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

**Solution:**

Given:n^{th}term of the series as:a

_{n}= (2n – 1)^{2}= 4n^{2}– 4n + 1So, the sum of n terms of the series: