Let n and k be positive integers such that n | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) The number of solutions of ${x_1} + {x_2} + ... + {x_k} = n$

= Coefficient of tn in $(t + {t^2} + {t^3} + ...)({t^2} + {t^3} + ...)$

Vedclass · Accepted Answer

$^m{C_{k - 1}}$

Vedclass · Answer

(a) The number of solutions of ${x_1} + {x_2} + ... + {x_k} = n$

= Coefficient of tn in $(t + {t^2} + {t^3} + ...)({t^2} + {t^3} + ...)$

                                                                      .....$({t^k} + {t^{k + 1}} + ....)$

= Coefficient of ${t^n}$in ${t^{1 + 2 + ... + k}}{(1 + t + {t^2} + ...)^k}$

But $1 + 2 + ... + k = \frac{1}{2}k(k + 1) = r$(say)

and $1 + t + {t^2} + .... = \frac{1}{{(1 - t)}}$

Thus number of required solutions

= Coefficient of ${t^{n - r}}$ in ${(1 - t)^{ - k}}$

=Coefficient of ${t^{n - r}}$ in

$[1 + {\,^k}{C_1}t + {\,^{k + 1}}{C_2}{t^2} + {\,^{k + 2}}{C_3}{t^3} + ...]$

$ = {\,^{k + n - r - 1}}{C_{k - 1}}$ ${x^2} - (p + \alpha )x + p\alpha + q\beta = 0$, (where

$m = k + n - r - 1$)

$ = k + n - 1 - \frac{1}{2}k(k + 1) = \frac{1}{2}[2k + 2n - 2 - {k^2} - k]$

$ = \frac{1}{2}(2n - {k^2} + k - 2)$

Let n and k be positive integers such that $n \ge \frac{{k(k + 1)}}{2}$. The number of solutions $({x_1},{x_2},....{x_k})$, ${x_1} \ge 1,{x_2} \ge 2,....{x_k} \ge k,$ all integers, satisfying ${x_1} + {x_2} + .... + {x_k} = n$, is

Similar Questions

If ${}^{21}{C_1} + 3.{}^{21}{C_3} + 5.{}^{21}{C_5} + ......19{}^{21}{C_{19}} + 21.{}^{21}{C_{21}} = k$ Then number of prime factors of $k$ is

For natural numbers $m,n$ ,if ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}y + {a_2}{y^2} + \ldots \;$ and $a_1= a_2=10,$ then $(m,n)$ =______.

The value of $\sum_{ r =0}^{6}\left({ }^{6} C _{ r }{ }^{-6} C _{6- r }\right)$ is equal to :

The sum of the coefficients in the expansion of ${(x + y)^n}$ is $4096$. The greatest coefficient in the expansion is

Let $\left( a + bx + cx ^2\right)^{10}=\sum \limits_{ i =0}^{20} p _{ i } x ^{ i }, a , b , c \in N$. If $p _1=20$ and $p _2=210$, then $2( a + b + c )$ is equal to