Notes: `3, -9/2, 27/4, -81/8` Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

Sunday, March 4, 2012

$\displaystyle{3},-\frac{{9}}{{2}},\frac{{27}}{{4}},-\frac{{81}}{{8}}$ Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

$\displaystyle{3},-\frac{{9}}{{2}},\frac{{27}}{{4}},-\frac{{81}}{{8}},\ldots.$

The terms of the above sequence can be written as,

$\displaystyle{3},{3}{\left(-\frac{{3}}{{2}}\right)},{3}{\left(-\frac{{3}}{{2}}\right)}{\left(-\frac{{3}}{{2}}\right)},{3}{\left(-\frac{{3}}{{2}}\right)}{\left(-\frac{{3}}{{2}}\right)}{\left(-\frac{{3}}{{2}}\right)},\ldots..$

So from the above we can see that the each consecutive term of the sequence is multiplied by a common ratio (-3/2).

So the nth term $\displaystyle{a}_{{n}}$ of the sequence can be written as $\displaystyle{3}\cdot{{\left(-\frac{{3}}{{2}}\right)}}^{{{n}-{1}}}$

In order to find the formula for the first n terms of the sequence, Let's first write down the first few sums of the sequence.

$\displaystyle{S}_{{1}}={3}$

$\displaystyle{S}_{{2}}={3}+{\left(-\frac{{9}}{{2}}\right)}=-\frac{{3}}{{2}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{2}}\right]}}}{{{1}-{\left(-\frac{{3}}{{2}}\right)}}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{2}}\right]}}}{{{1}+\frac{{3}}{{2}}}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{2}}\right]}$

$\displaystyle{S}_{{3}}={3}+{\left(-\frac{{9}}{{2}}\right)}+\frac{{27}}{{4}}=\frac{{21}}{{4}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{3}}\right]}}}{{{1}-{\left(-\frac{{3}}{{2}}\right)}}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{3}}\right]}}}{{{1}+\frac{{3}}{{2}}}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{3}}\right]}$

$\displaystyle{S}_{{4}}={3}+{\left(-\frac{{9}}{{2}}\right)}+\frac{{27}}{{4}}+{\left(-\frac{{81}}{{8}}\right)}=-\frac{{39}}{{8}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{4}}\right]}}}{{{1}-{\left(-\frac{{3}}{{2}}\right)}}}=\frac{{{3}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{4}}\right]}}}{{{1}+\frac{{3}}{{2}}}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{4}}\right]}$

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

$\displaystyle{S}_{{k}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}\right]}$

Let's check the above formula for n=1,

$\displaystyle{S}_{{1}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{1}}\right]}=\frac{{6}}{{5}}{\left({1}+\frac{{3}}{{2}}\right)}=\frac{{6}}{{5}}{\left(\frac{{5}}{{2}}\right)}={3}$

So the formula is verified for n=1.

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

$\displaystyle{S}_{{{k}+{1}}}={3}+{\left(-\frac{{9}}{{2}}\right)}+\frac{{27}}{{4}}+{\left(-\frac{{81}}{{8}}\right)}+\ldots\ldots+{3}{{\left(-\frac{{3}}{{2}}\right)}}^{{{k}-{1}}}+{3}{{\left(-\frac{{3}}{{2}}\right)}}^{{{k}+{1}-{1}}}$

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

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}\right]}+{3}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}-\frac{{6}}{{5}}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}+{3}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}+{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}{\left({3}-\frac{{6}}{{5}}\right)}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}+{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}{\left(\frac{{{15}-{6}}}{{5}}\right)}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}+\frac{{9}}{{5}}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}{\left[{1}+\frac{{9}}{{6}}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}\right]}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}{\left[{1}+\frac{{3}}{{2}}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}\right]}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}{\left[{1}-{\left(-\frac{{3}}{{2}}\right)}{{\left(-\frac{{3}}{{2}}\right)}}^{{k}}\right]}$

$\displaystyle{S}_{{{k}+{1}}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{{k}+{1}}}\right]}$

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

Hence the formula for the sum of the n terms of the sequence is,

$\displaystyle{S}_{{n}}=\frac{{6}}{{5}}{\left[{1}-{{\left(-\frac{{3}}{{2}}\right)}}^{{n}}\right]}$

Notes

Sunday, March 4, 2012

$\displaystyle{3},-\frac{{9}}{{2}},\frac{{27}}{{4}},-\frac{{81}}{{8}}$ Use mathematical induction to find a formula for the sum of the first n terms of the sequence.

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