Question

In this sequence, term number 1 has a value of 6,

Answer 1

Answer:

2n² + 4

Step-by-step explanation:

Let, $S_{n} = 6 +12 +22 +36 +54 + ........ +t_{n-1} +t_{n}$ ...... (1)

Now, sift the right hand side by one term and subtract from original equation (1).

Hence, we get  

($S_{n} -S_{n} =6+ [(12-6) + (22-12) + (36-22) + (54-36) + ....... ] - t_{n}$

⇒ tₙ = 6 + [ 6 + 10 + 14 + 18 + ........ up to (n-1)th term]  

Now, the sum within the bracket is an A.P. sum.

Hence, tₙ = 6+ [$\frac{n-1}{2}(2*6+(n-2)*4)$]

               = 6+ $\frac{n-1}{2} (4n+4)$

               = 6+ 2(n²-1)

                = 2n² + 4

Therefore, the general term 2n² + 4 represent the sequence. (Answer)