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Question Number 23953 by NECx last updated on 10/Nov/17
find the nth term of the sequence 4, 9, 16, 45, 76 please help
$${find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{45},\:\mathrm{76} \\ $$$$ \\ $$$${please}\:{help} \\ $$
Commented by prakash jain last updated on 10/Nov/17
T_n =((4(n−2)(n−3)(n−4)(n−5))/(1∙2∙3∙4)) −((9(n−1)(n−3)(n−4)(n−5))/(1∙2∙3)) +((16(n−1)(n−2)(n−4)(n−5))/(1∙2∙1∙2)) −((45(n−1)(n−2)(n−3)(n−5))/(1∙2∙3)) +((76(n−1)(n−2)(n−3)(n−4))/(1∙2∙3∙4)) +k(n−1)(n−2)(n−3)(n−4)(n−5) k is any value.
$${T}_{{n}} =\frac{\mathrm{4}\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:−\frac{\mathrm{9}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{16}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{1}\centerdot\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:−\frac{\mathrm{45}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{5}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{76}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{k}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right) \\ $$$$ \\ $$$${k}\:\mathrm{is}\:\mathrm{any}\:\mathrm{value}. \\ $$
Commented by NECx last updated on 10/Nov/17
hmmmm
$${hmmmm} \\ $$
Commented by prakash jain last updated on 10/Nov/17
It satisfies the given conditions n=1 to 5 for any value of k. In general the point is that there are infinite number of formula available for a given number of finite terms.
$$\mathrm{It}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{given}\:\mathrm{conditions} \\ $$$${n}=\mathrm{1}\:\mathrm{to}\:\mathrm{5}\:\mathrm{for}\:\mathrm{any}\:\mathrm{value}\:\mathrm{of}\:{k}. \\ $$$$\mathrm{In}\:\mathrm{general}\:\mathrm{the}\:\mathrm{point}\:\mathrm{is}\:\mathrm{that}\:\mathrm{there} \\ $$$$\mathrm{are}\:\mathrm{infinite}\:\mathrm{number}\:\mathrm{of}\:\mathrm{formula} \\ $$$$\mathrm{available}\:\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{finite}\:\mathrm{terms}. \\ $$
Commented by NECx last updated on 10/Nov/17
thankz a lot.
$${thankz}\:{a}\:{lot}. \\ $$

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