Question

Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2 +9n+13, for every positive integer n ≥ 2. If tk=103, then k equals

Answer 1

Correct Answer: 24

Step 1: Sum Formula for First \( n \) Terms 

We are given the sum of the first \( n \) terms of a sequence:

\[ S_n = t_1 + t_2 + \cdots + t_n = 2n^2 + 9n + 13 \tag{1} \]

Sum of the first \( n - 1 \) terms:

\[ S_{n-1} = t_1 + t_2 + \cdots + t_{n-1} = 2(n-1)^2 + 9(n-1) + 13 \tag{2} \]

Step 2: Find the General Term \( t_n \)

Using the identity \( t_n = S_n - S_{n-1} \), subtract (2) from (1):

\[ t_n = \left( 2n^2 + 9n + 13 \right) - \left( 2(n - 1)^2 + 9(n - 1) + 13 \right) \]

Expand and simplify the second expression:

\[ 2(n - 1)^2 = 2(n^2 - 2n + 1) = 2n^2 - 4n + 2 \]

9(n - 1) = 9n - 9 \]

So:

\[ t_n = \left( 2n^2 + 9n + 13 \right) - \left( 2n^2 - 4n + 2 + 9n - 9 + 13 \right) = (2n^2 + 9n + 13) - (2n^2 + 5n + 6) = 4n + 7 \]

Step 3: Use Given Value \( t_k = 103 \)

We are told \( t_k = 103 \), so:

\[ 4k + 7 = 103 \Rightarrow 4k = 96 \Rightarrow k = \frac{96}{4} = \boxed{24} \]

Final Answer:

\[ \boxed{k = 24} \]