In this sequence, term number 1 has a value of 6,term number 2 has a value of 12, term number 3 hasa value of 22, and so on:6, 12, 22, 36, 54, ...If n represents the term number, demonstrate thatthe rule 2n + 4 will generate the sequence.
Accepted Solution
A:
Answer: 2n² + 4 Step-by-step explanation:Let, [tex]S_{n} = 6 +12 +22 +36 +54 + ........ +t_{n-1} +t_{n}[/tex] ...... (1)
Now, sift the right hand side by one term and subtract from original equation (1).
Hence, we get ([tex]S_{n} -S_{n} =6+ [(12-6) + (22-12) + (36-22) + (54-36) + ....... ] - t_{n}[/tex]
⇒ 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+ [[tex]\frac{n-1}{2}(2*6+(n-2)*4)[/tex]]
= 6+ [tex]\frac{n-1}{2} (4n+4)[/tex]
= 6+ 2(n²-1)
= 2n² + 4
Therefore, the general term 2n² + 4 represent the sequence. (Answer)