Sn in terms of n

Let: ${ S }_{ n }=2n-3+{ S }_{ n-2 }$ for $n>1$

Find a rule for ${ S }_{ n }$ in terms of $n$ if ${ S }_{ 1 }=1$ , where $n\in Odd\quad numbers$

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

If $S_n$ is a polynomial, the degree of $S_n-S_{n-2}=2n-3$ is one less than the degree of $S_n$ , making $S_n$ quadratic.

Since $\dfrac{1}{2}n^2-\dfrac{1}{2}(n-2)^2-\bigg(\dfrac{1}{2}n-\dfrac{1}{2}(n-2)\bigg)=2n-3$ , and $S_1=1$ , we have that $S_n=\dfrac{1}{2}n^2-\dfrac{1}{2}n+1$ will work for all positive, odd $n$ .

Miles Koumouris - 3 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$