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Question

Question: Find the number of negative terms in the sequence : \({x_n} = \dfrac{{{}_{}^{n + 4}{P_4}}}{{{P_{n ...

Find the number of negative terms in the sequence :
xn=n+4P4Pn+21434Pn{x_n} = \dfrac{{{}_{}^{n + 4}{P_4}}}{{{P_{n + 2}}}} - \dfrac{{143}}{{4{P_n}}}

Explanation

Solution

In the given question the term Pn{P_n} and Pn+2{P_{n + 2}} stands for n+2Pn+2{}^{n + 2}{P_{n + 2}} and nPn{}^n{P_n} respectively. Therefore , we know that for any permutation expression nPn=P!{}^n{P_n} = P!. Now solve it using a simple factorial rule.

Complete step by step answer:
We can write the given equation also as : -
xn=n+4P4n+2Pn+21434nPn{x_n} = \dfrac{{{}_{}^{n + 4}{P_4}}}{{{}^{n + 2}{P_{n + 2}}}} - \dfrac{{143}}{{4{}^n{P_n}}}
Now using the formula of permutation
nPr=n!(nr)!{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}
Substituting the values we get,
xn=(n+4)!(n44)!(n+2)!1434n!{x_n} = \dfrac{{\dfrac{{(n + 4)!}}{{(n - 4 - 4)!}}}}{{(n + 2)!}} - \dfrac{{143}}{{4n!}}
xn=(n+4)!n!(n+2)!1434n!\Rightarrow {x_n} = \dfrac{{\dfrac{{(n + 4)!}}{{n!}}}}{{(n + 2)!}} - \dfrac{{143}}{{4n!}}
Further solving the factorial in the numerator we get,
xn=(n+4)(n+3)(n+2)(n+1)n!n!(n+2)!1434.n!{x_n} = \dfrac{{\dfrac{{(n + 4)(n + 3)(n + 2)(n + 1)n!}}{{n!}}}}{{(n + 2)!}} - \dfrac{{143}}{{4.n!}}
xn=(n+4)(n+3)(n+2)(n+1)(n+2)!1434.n!{x_n} = \dfrac{{(n + 4)(n + 3)(n + 2)(n + 1)}}{{(n + 2)!}} - \dfrac{{143}}{{4.n!}}

Further solving the factorial in the denominator we get,
xn=(n+4)(n+3)(n+2)(n+1)(n+2)(n+1)n!1434.n!{x_n} = \dfrac{{(n + 4)(n + 3)(n + 2)(n + 1)}}{{(n + 2)(n + 1)n!}} - \dfrac{{143}}{{4.n!}}
xn=(n+4)(n+3)n!1434.n!\Rightarrow {x_n} = \dfrac{{(n + 4)(n + 3)}}{{n!}} - \dfrac{{143}}{{4.n!}}
On taking L . C . M , we get
xn=(4n2+28n95)4.n!{x_n} = \dfrac{{(4{n^2} + 28n - 95)}}{{4.n!}}
Now , since we have to find negative terms in xn{x_n} , so xn{x_n} will be less zero .
Therefore , xn=(4n2+28n95)4.n!<0{x_n} = \dfrac{{(4{n^2} + 28n - 95)}}{{4.n!}} < 0
xn=(4n2+28n95)<0{x_n} = (4{n^2} + 28n - 95) < 0 , for solving the value of nn we factorize the expression.
The given could be satisfied for the values n=1,2n = 1,2.

Therefore , on putting values of nn in equation , we get
x1=634\therefore {x_1} = - \dfrac{{63}}{4} and x2=238{x_2} = - \dfrac{{23}}{8} .

So, there are two negative values for the given sequence.

Note: If nn and rr are positive integer such that 1rn1 \leqslant r \leqslant n , then the number of all permutation of nn distinct things , taken rr at a time is denoted by symbol P(n,r)P(n,r) or nPr{}^n{P_r}. It should be noted that in permutations , the order of arrangement is taken into account when the order is changed , a different permutation is obtained. In permutations , if there are two jobs such that one of them can be completed in nn ways , and when it has been completed in any one of these nn ways , second job can be completed in rr, then the two jobs succession can be completed in n×rn \times r ways.