Find nth term of the expression: 3,8,15,24,35

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on Mar 30, 2025 at 5:52 am



MathBot Answer:

3,8,15,24,35,...3,8,15,24,35,...This infinite sequence is a quadratic sequence. The nth term rule for the sequence is an=n(n+2)a_n=n \left(n + 2\right), where n>0n>0. The recursive formula is an=an1+2n+1a_n=a_{n-1} + 2 n + 1, where n>1n>1 and a1=3a_{1}=3.


38152435+5+7+9+11+2+2+2\begin{matrix}3&&8&&15&&24&&35\\[6pt]&+5&&+7&&+9&&+11\\[6pt]&&+2&&+2&&+2\\[6pt]\end{matrix}

Explicit Formula

Since there are 2 rows of differences, the formula for the sequence can be written as a polynomial with degree 2, where nn is the term number and (x0,x1,x2)(x_{0}, x_{1}, x_{2}) are the coefficients: an=n2x2+nx1+x0a_n=n^{2} x_{2} + n x_{1} + x_{0}

Using the first 3 terms in the sequence, create and solve the system of equations for (x0,x1,x2)(x_{0}, x_{1}, x_{2}): 3=12x2+1x1+x08=22x2+2x1+x015=32x2+3x1+x0x0+x1+x2=3x0+2x1+4x2=8x0+3x1+9x2=15\begin{aligned} 3 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ 8 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ 15 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = 3\\x_{0} + 2 x_{1} + 4 x_{2} = 8\\x_{0} + 3 x_{1} + 9 x_{2} = 15 \end{aligned} (x0,x1,x2)=(0, 2, 1) \Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( 0, \ 2, \ 1\right)

The nth term rule is:an=n2x2+nx1+x0=n2(1)+n(2)+(0)=n(n+2)\begin{aligned} a_n&=n^{2} x_{2} + n x_{1} + x_{0} \\ &=n^{2} \left(1\right) + n \left(2\right) + \left(0\right) \\ &=n \left(n + 2\right) \end{aligned}

Recursive Formula

Since there are 2 rows of differences, the formula for the sequence can be written as the sum of an1a_{n-1} and polynomial with degree 1, where nn is the term number and (x0,x1)(x_{0}, x_{1}) are the coefficients: an=an1+nx1+x0a_n=a_{n-1} + n x_{1} + x_{0}

Using the first 3 terms in the sequence, create and solve the system of equations for (x0,x1)(x_{0}, x_{1}): 8=3+2x1+x015=8+3x1+x0x0+2x1=5x0+3x1=7\begin{aligned} 8 &= 3 + 2 x_{1} + x_{0} \\ 15 &= 8 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = 5\\x_{0} + 3 x_{1} = 7 \end{aligned} (x0,x1)=(1, 2) \Rightarrow \quad (x_{0}, x_{1})=\left( 1, \ 2\right)

The nth term rule is:an=an1+nx1+x0=an1+n(2)+(1)=an1+2n+1\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(2\right) + \left(1\right) \\ &=a_{n-1} + 2 n + 1 \end{aligned}