In series 7+12+20+31+......................................... Then 1)tn. 2) sn

To tackle the series you've presented, let's first identify the pattern in the sequence: 7, 12, 20, 31, and so on. We can break this down step by step to find both the general term (tn) and the sum of the first n terms (sn).

Identifying the Pattern

To find the pattern, we can look at the differences between consecutive terms:

• 12 - 7 = 5
• 20 - 12 = 8
• 31 - 20 = 11

The differences are 5, 8, and 11. If we examine these differences further, we see that they increase by 3 each time:

• 8 - 5 = 3
• 11 - 8 = 3

This suggests that the nth term of the series can be expressed as a quadratic function, since the second difference is constant.

Finding the General Term (tn)

We can express the nth term of the series as:

tn = an² + bn + c

To find the coefficients a, b, and c, we can set up a system of equations using the first few terms:

• For n=1: a(1)² + b(1) + c = 7
• For n=2: a(2)² + b(2) + c = 12
• For n=3: a(3)² + b(3) + c = 20

This gives us the following equations:

• 1a + 1b + c = 7
• 4a + 2b + c = 12
• 9a + 3b + c = 20

By solving this system, we can find the values of a, b, and c. Subtracting the first equation from the second and the second from the third helps eliminate c:

• (4a + 2b + c) - (1a + 1b + c) = 12 - 7 → 3a + b = 5
• (9a + 3b + c) - (4a + 2b + c) = 20 - 12 → 5a + b = 8

Now we have a simpler system:

• 3a + b = 5
• 5a + b = 8

Subtracting these two equations gives:

• (5a + b) - (3a + b) = 8 - 5 → 2a = 3 → a = 1.5

Substituting a back into one of the equations to find b:

• 3(1.5) + b = 5 → 4.5 + b = 5 → b = 0.5

Now substituting a and b into one of the original equations to find c:

• 1.5 + 0.5 + c = 7 → c = 5

Thus, the general term is:

tn = 1.5n² + 0.5n + 5

Calculating the Sum of the First n Terms (sn)

To find the sum of the first n terms, we can use the formula for the sum of a quadratic sequence:

sn = n/6(2a + (n - 1)d + (n - 1)(n - 2)c)

Where d is the first difference and c is the second difference. In our case, d = 3 and c = 3. Plugging in the values:

• sn = n/6(2(1.5) + (n - 1)(3) + (n - 1)(n - 2)(3))

This can be simplified further, but for practical purposes, you can calculate sn for specific values of n using the derived formula for tn.

Example Calculation

Let’s say we want to find the sum of the first 4 terms:

• t1 = 7
• t2 = 12
• t3 = 20
• t4 = 31

So, sn = 7 + 12 + 20 + 31 = 70.

In summary, the general term of the series is given by tn = 1.5n² + 0.5n + 5, and the sum of the first n terms can be calculated using the derived formulas. This approach not only helps in understanding the series but also equips you with the tools to analyze similar sequences in the future.