### Sequences and Series Exercise 4

## Question:

1/(1×2)+1/(2×3)+1/(3×4)+….

## Answer:

Step 1: Rewrite the expression as 1/2 + 1/6 + 1/12 + ….

Step 2: Factor out the common denominator 1/2(1 + 1/3 + 1/6 + ….)

Step 3: Rewrite the expression as a geometric series 1/2(1 + r + r2 + r3 + ….)

Step 4: Calculate the sum of the series 1/2(1/(1-r))

## Question:

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

## Answer:

Step 1: The general term of the series can be written as Tn = n(n+1)(n+2).

Step 2: The sum of the series can be written as S = T1 + T2 + T3 + … + Tn.

Step 3: Substitute the value of Tn in the equation.

Step 4: S = 1(2)(3) + 2(3)(4) + 3(4)(5) + … + n(n+1)(n+2).

Step 5: Simplify the equation by factoring out the common terms.

Step 6: S = n(n+1)(n+2)(1 + 2 + 3 + … + n).

Step 7: Use the formula for the sum of an arithmetic progression to simplify the equation.

Step 8: S = n(n+1)(n+2) × (n(n+1))/2.

Step 9: Simplify the equation by factoring out the common terms.

Step 10: S = (n(n+1)(n+2))/2 × (n(n+1)).

Step 11: The sum of the series up to n terms is S = (n(n+1)(n+2))/2 × (n(n+1)).

## Question:

Find nth term of the series 3×1^2,5×2^2,7×3^2,…..

## Answer:

Step 1: Identify the pattern.

The pattern is 3×1^2, 5×2^2, 7×3^2, 9×4^2, 11×5^2, …

Step 2: Determine the common difference.

The common difference is 2.

Step 3: Find the nth term.

The nth term is (2n+1)×n^2.

## Question:

Find the sum of the series : (5^2+6^2+7^2+…+20^2)

## Answer:

Answer: Step 1: Find the first and last terms of the series: First term: 5^2 = 25 Last term: 20^2 = 400

Step 2: Calculate the number of terms in the series: Number of terms = Last term - First term + 1 = 400 - 25 + 1 = 376

Step 3: Calculate the sum of the series: Sum of the series = n/2[2a + (n-1)d] = 376/2[2(25) + (376-1)(1)] = 376/2[50 + 375] = 376/825[425] = 17,376

## Question:

The sum of n terms of 1^2+(1^2+2^2)+(1^2+2^2+3^2)+….. A : n(n+a)(2n+1)/6 B : n(n+1)(2n−1)/6 C : n(n+1)2(n+2)1/12 D : (n^2(n+1)^2)1/12

## Answer:

Answer: A : n(n+1)(2n+1)/6

## Question:

Find the sum to n terms of the series whose nth term is n(n+4).

## Answer:

Answer:

Step 1: Find the formula for the sum of the series. The formula for the sum of the series is S = (n/2)(2a + (n - 1)d), where a is the first term and d is the common difference.

Step 2: Substitute the values of a and d in the formula. In this case, a = n and d = 4, so the formula becomes S = (n/2)(2n + (n - 1)4).

Step 3: Simplify the formula to get the final answer. S = (n/2)(2n + 4n - 4) = (n/2)(6n - 4) = 3n^2 - 2n

## Question:

Find the sum of n terms of the series whose nth term is : n^2+2^n

## Answer:

Solution:

Step 1: Find the first term (a1)

a1 = 12 + 2^1

a1 = 14

Step 2: Find the common difference (d)

d = n2 + 2n - (n-1)2 + 2n-1

d = 2n

Step 3: Find the sum of n terms

Sn = n/2 [2a1 + (n-1)d]

Sn = n/2 [2(14) + (n-1)(2n)]

Sn = n/2 [28 + 2n2 - 2n]

Sn = n2 + n - 14

## Question:

Find the sum of n terms of the series whose nth term is : (2n−1)^2

## Answer:

Answer: Step 1: Write down the formula for the sum of n terms of a series: Sn = n/2[2a + (n-1)d]

Step 2: Substitute the values of a (first term) and d (common difference) in the formula: Sn = n/2[2(2*1-1)^2 + (n-1)(2-1)^2]

Step 3: Simplify the equation: Sn = n/2[4 + (n-1)1]

Step 4: Solve for Sn: Sn = n/2[n+3]

Hence, the sum of n terms of the series is Sn = n/2[n+3].

## Question:

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

## Answer:

Step 1: Find the common difference between the terms.

The common difference between the terms is 3.

Step 2: Find the last term of the series.

Let n be the number of terms in the series.

The last term of the series will be (n-1) × (3 + 3 × (n-1)) = 3n2-3n+3.

Step 3: Find the sum of the series.

The sum of the series is given by the formula:

Sn = [n/2] × (a1 + an)

where a1 is the first term and an is the last term of the series.

In this case, a1 = 3 and an = 3n2-3n+3.

Therefore, the sum of the series is given by:

Sn = [n/2] × (3 + 3n2-3n+3)

Sn = [n/2] × (3n2-3n+6)

Sn = [n2/2] × (3n-3) + [n/2] × 6

Sn = (3n3-3n2)/2 + 3n

Therefore, the sum of the series 3×8+6×11+9×14+…. to n terms is (3n3-3n2)/2 + 3n.

## Question:

The sum of the series 1+2×3+3×5+4×7+….. A : 915 B : 946 C : 945 D : 916

## Answer:

- The series can be written as 1 + 2(3) + 3(5) + 4(7) + …
- The next two terms are 5(9) and 6(11).
- The sum of the series can be calculated by adding all the terms together: 1 + 2(3) + 3(5) + 4(7) + 5(9) + 6(11) = 945

Answer: C. 945