`1, 5, 9, 13...` Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

Saturday, September 27, 2014

$\displaystyle{1},{5},{9},{13}\ldots$ Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

$\displaystyle{1},{5},{9},{13},\ldots\ldots$

Now let's denote the first term of the above sequence by $\displaystyle{a}_{{1}}$ and k'th term by $\displaystyle{a}_{{k}}$ and let's write down the first few sums of the sequence.

$\displaystyle{S}_{{1}}={1}$

$\displaystyle{S}_{{2}}={1}+{5}={6}=\frac{{{2}{\left({1}+{5}\right)}}}{{2}}=\frac{{{2}{\left({a}_{{1}}+{a}_{{2}}\right)}}}{{2}}$

$\displaystyle{S}_{{3}}={1}+{5}+{9}={15}=\frac{{{3}{\left({1}+{9}\right)}}}{{2}}=\frac{{{3}{\left({a}_{{1}}+{a}_{{3}}\right)}}}{{2}}$

$\displaystyle{S}_{{4}}={1}+{5}+{9}+{13}={28}=\frac{{{4}{\left({1}+{13}\right)}}}{{2}}=\frac{{{4}{\left({a}_{{1}}+{a}_{{4}}\right)}}}{{2}}$

From the above sequence it appears that the formula for the sum of the k terms is,

$\displaystyle{S}_{{k}}=\frac{{{k}{\left({1}+{a}_{{k}}\right)}}}{{2}}$

Now the difference between the consecutive terms of the series is 4, so we can write down ,

$\displaystyle{a}_{{k}}={1}+{\left({k}-{1}\right)}{4}={1}+{4}{k}-{4}={4}{k}-{3}$

$\displaystyle{a}_{{{k}+{1}}}={4}{\left({k}+{1}\right)}-{3}={4}{k}+{4}-{3}={4}{k}+{1}$

$\displaystyle{S}_{{k}}=\frac{{{k}{\left({1}+{4}{k}-{3}\right)}}}{{2}}=\frac{{{k}{\left({4}{k}-{2}\right)}}}{{2}}$

$\displaystyle{S}_{{k}}={k}{\left({2}{k}-{1}\right)}$

Now...

$\displaystyle{1},{5},{9},{13},\ldots\ldots$

Now let's denote the first term of the above sequence by $\displaystyle{a}_{{1}}$ and k'th term by $\displaystyle{a}_{{k}}$ and let's write down the first few sums of the sequence.

$\displaystyle{S}_{{1}}={1}$

$\displaystyle{S}_{{2}}={1}+{5}={6}=\frac{{{2}{\left({1}+{5}\right)}}}{{2}}=\frac{{{2}{\left({a}_{{1}}+{a}_{{2}}\right)}}}{{2}}$

$\displaystyle{S}_{{3}}={1}+{5}+{9}={15}=\frac{{{3}{\left({1}+{9}\right)}}}{{2}}=\frac{{{3}{\left({a}_{{1}}+{a}_{{3}}\right)}}}{{2}}$

$\displaystyle{S}_{{4}}={1}+{5}+{9}+{13}={28}=\frac{{{4}{\left({1}+{13}\right)}}}{{2}}=\frac{{{4}{\left({a}_{{1}}+{a}_{{4}}\right)}}}{{2}}$

From the above sequence it appears that the formula for the sum of the k terms is,

$\displaystyle{S}_{{k}}=\frac{{{k}{\left({1}+{a}_{{k}}\right)}}}{{2}}$

Now the difference between the consecutive terms of the series is 4, so we can write down ,

$\displaystyle{a}_{{k}}={1}+{\left({k}-{1}\right)}{4}={1}+{4}{k}-{4}={4}{k}-{3}$

$\displaystyle{a}_{{{k}+{1}}}={4}{\left({k}+{1}\right)}-{3}={4}{k}+{4}-{3}={4}{k}+{1}$

$\displaystyle{S}_{{k}}=\frac{{{k}{\left({1}+{4}{k}-{3}\right)}}}{{2}}=\frac{{{k}{\left({4}{k}-{2}\right)}}}{{2}}$

$\displaystyle{S}_{{k}}={k}{\left({2}{k}-{1}\right)}$

Now we can verify that it is valid for n=1,

Plug in k=1 to verify,

$\displaystyle{S}_{{1}}={1}{\left({2}\cdot{1}-{1}\right)}={1}$

Let's assume that the formula is valid for n=k and now we have to show that it is valid for n=k+1

$\displaystyle{S}_{{{k}+{1}}}={1}+{5}+{9}+{13}\ldots\ldots..+{a}_{{k}}+{a}_{{{k}+{1}}}$

$\displaystyle{S}_{{{k}+{1}}}={S}_{{k}}+{a}_{{{k}+{1}}}$

$\displaystyle{S}_{{{k}+{1}}}={k}{\left({2}{k}-{1}\right)}+{4}{k}+{1}$

$\displaystyle{S}_{{{k}+{1}}}={2}{{k}}^{{2}}-{k}+{4}{k}+{1}$

$\displaystyle{S}_{{{k}+{1}}}={2}{{k}}^{{2}}+{3}{k}+{1}$

$\displaystyle{S}_{{{k}+{1}}}={2}{{k}}^{{2}}+{2}{k}+{k}+{1}$

$\displaystyle{S}_{{{k}+{1}}}={2}{k}{\left({k}+{1}\right)}+{1}{\left({k}+{1}\right)}$

$\displaystyle{S}_{{{k}+{1}}}={\left({k}+{1}\right)}{\left({2}{k}+{1}\right)}$

$\displaystyle{S}_{{{k}+{1}}}={\left({k}+{1}\right)}{\left[{2}{\left({k}+{1}\right)}-{1}\right]}$

So the formula is valid for n=k+1 also,

So the formula can be written as ,

$\displaystyle{S}_{{n}}={n}{\left({2}{n}-{1}\right)}$

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