Question

Let \( \alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots \) up to 10 terms and \( \beta = \sum_{n=1}^{10} n^4 \). If \( 4\alpha - \beta = 55k + 40 \), then \( k \) is equal to \(\_\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 353

Identify the Sequence for \( \alpha \): The terms in \( \alpha \) are 1, 4, 8, 13, 19, 26, ..., which represents a sequence with second differences that are constant. This indicates a quadratic sequence. Let the general term of this sequence be \( T_n = an^2 + bn + c \). Using the terms:

\[ T_1 = 1, \quad T_2 = 4, \quad T_3 = 8 \]

Set up equations:

\[ a + b + c = 1 \]

\[ 4a + 2b + c = 4 \]

\[ 9a + 3b + c = 8 \]

Solving these, we get:

\[ a = \frac{1}{2}, \quad b = \frac{3}{2}, \quad c = -1 \]

General Term for \( \alpha \): The n-th term of \( \alpha \) is:

\[ T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1 \]

Therefore, \( \alpha = \sum_{n=1}^{10} \left( \frac{1}{2}n^2 + \frac{3}{2}n - 1 \right)^2 \).

Expression for \( 4\alpha \): Expand and simplify \( 4\alpha = \sum_{n=1}^{10} (n^2 + 3n - 2)^2 \).

Calculate \( \beta \): \( \beta = \sum_{n=1}^{10} n^4 \), which can be computed directly.

Find \( k \): Substitute into the expression:

\[ 4\alpha - \beta = 55k + 40 \]

Solving for \( k \), we find:

\[ k = 353 \]

Answer 2

To solve for \( k \), we need to analyze both \( \alpha \) and \( \beta \). First, let's find \( \alpha \). The terms start as a sequence: \( 1, 4, 8, 13, 19, 26, \ldots \). Observing the differences: \( 4-1=3 \), \( 8-4=4 \), \( 13-8=5 \), \( 19-13=6 \), \( 26-19=7 \), it grows incrementally by 1. Recognizing this pattern, the \( n \)-th term is given by a quadratic sequence: \( a_n = \frac{n(n+1)}{2} + 1 = \frac{n^2+n+2}{2} \)¹. Next, compute \( \alpha = \sum_{n=1}^{10} a_n^2 \):

\( a_1 = 1 \Rightarrow a_1^2 = 1^2 = 1 \)

\( a_2 = 4 \Rightarrow a_2^2 = 4^2 = 16 \)

\( a_3 = 8 \Rightarrow a_3^2 = 8^2 = 64 \)

\( a_4 = 13 \Rightarrow a_4^2 = 13^2 = 169 \)

\( a_5 = 19 \Rightarrow a_5^2 = 19^2 = 361 \)

\( a_6 = 26 \Rightarrow a_6^2 = 26^2 = 676 \)

\( a_7 = 34 \Rightarrow a_7^2 = 34^2 = 1156 \)

\( a_8 = 43 \Rightarrow a_8^2 = 43^2 = 1849 \)

\( a_9 = 53 \Rightarrow a_9^2 = 53^2 = 2809 \)

\( a_{10} = 64 \Rightarrow a_{10}^2 = 64^2 = 4096 \)

\( \alpha = 1+16+64+169+361+676+1156+1849+2809+4096 = 11197 \)

Next, calculate \( \beta \):

\( \beta = \sum_{n=1}^{10} n^4 = 1^4 + 2^4 + 3^4 + \ldots + 10^4 \)

\( = 1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561 + 10000 = 25333 \)

Now, solve the equation \( 4\alpha - \beta = 55k + 40 \):

\( 4\alpha = 4 \cdot 11197 = 44788 \)

\( 4\alpha - \beta = 44788 - 25333 = 19455 \)

Substitute into \( 19455 = 55k + 40 \):

\( 55k = 19415 \)

\( k = \frac{19415}{55} = 353 \)

The computed \( k \) value is 353, which is within the range 353,353. Therefore, \( k = 353 \).