Exploring some more Progressions - Alternative Shorter Method for tn-advanced

Find the general term $t_n$ for the sequence: $4, 11, 22, 37, 56, \dots$

Step 1: Find the differences. Terms: $4, 11, 22, 37, 56$ First differences: $11-4=7, 22-11=11, 37-22=15, 56-37=19$ Second differences: $11-7=4, 15-11=4, 19-15=4$ Since the second difference is constant ($d_2 = 4$), the sequence is quadratic: $t_n = an^2 + bn + c$.

Step 2: Calculate coefficients using the shorter method: $2a = 4 \implies a = 2$ $3a + b = 7 \implies 3(2) + b = 7 \implies 6 + b = 7 \implies b = 1$ $a + b + c = 4 \implies 2 + 1 + c = 4 \implies 3 + c = 4 \implies c = 1$

Step 3: Write the final formula. $t_n = 2n^2 + n + 1$

We first verified that the second difference is constant. We then used the relations $2a = d_2$, $3a + b = t_2 - t_1$, and $a + b + c = t_1$ to find the coefficients of the quadratic equation quickly.

Find the $10^{th}$ term of the sequence whose first differences are $5, 8, 11, 14, \dots$ and the first term $t_1 = 2$.

Step 1: Identify differences. First differences ($d_1$): $5, 8, 11, 14$ Second differences ($d_2$): $8-5=3, 11-8=3$ Constant second difference $d_2 = 3$. First term $t_1 = 2$. First first-difference $\Delta_1 = 5$.

Step 2: Use the formula $t_n = t_1 + (n-1)\Delta_1 + \frac{(n-1)(n-2)}{2}\Delta_2$. For $n=10$: $t_{10} = 2 + (10-1)(5) + \frac{(10-1)(10-2)}{2}(3)$ $t_{10} = 2 + (9)(5) + \frac{9 \times 8}{2}(3)$ $t_{10} = 2 + 45 + 36 \times 3$ $t_{10} = 47 + 108$ $t_{10} = 155$

Instead of finding the general formula $t_n$ first, we applied the expanded difference formula directly for $n = 10$ to save time.

Calculate the first difference for the terms $23$ and $50$ vertically.

$\begin{array}{r} 50 \\ -23 \\ \hline 27 \end{array}$

The difference between the two consecutive terms $t_k$ and $t_{k+1}$ gives the value for the first difference row.