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Find the coefficient of x83 in (x5+ x8+ x11+ x14+ x17)10 ?

asked in Combinatory by Veteran (11.9k points)  
retagged by | 368 views
can somebody provide me exact answer........

Debashish Deka's ans is correct! What else you need?

2 Answers

+11 votes
Best answer

$\begin{align*} &\left [ {\color{red}{\bf x^{83}}} \right ] : \left [ \left ( x^5+x^8 + x^{11} + x^{14} + x^{17}\right )^{10} \right ] \\ &\left [ {\color{red}{\bf x^{83}}} \right ] :\left [ x^{50}\left ( 1+x^3 + x^{6} + x^{9} + x^{12} \right )^{10} \right ] \\ &\left [ {\color{red}{\bf x^{33}}} \right ] :\left [\left ( 1+x^3 + x^{6} + x^{9} + x^{12} \right )^{10} \right ] \\ &\left [ {\color{red}{\bf x^{33}}} \right ] :\left [ \left \{ 1+(x^3)^1 + (x^{3})^2 + (x^{3})^3 + (x^{3})^4 \right \}^{10} \right ] \\ &\left [ {\color{red}{\bf x^{33}}} \right ] : \left [ \left \{ \frac{1-(x^3)^{4+1}}{1-(x^3)} \right \}^{10} \right ] \\ &\left [ {\color{red}{\bf x^{33}}} \right ] : \left [ \left ( 1-x^{15} \right )^{10}.\frac{1}{\left ( 1-x^3 \right )^{10}} \right ] \\ &\left [ {\color{red}{\bf x^{33}}} \right ] : \left [ \sum_{r=0}^{10}\binom{10}{r}\left ( -x^{15} \right )^r \; . \; \sum_{k=0}^{\infty}\binom{10+k-1}{k} . (x^3)^k \right ] \\ \end{align*}$

 

Now,

$\begin{align*} &\left [ {\color{red}{\bf x^{33}}} \right ] : \left [ \sum_{r=0}^{10}\binom{10}{r}\left ( -x^{15} \right )^r \; . \; \sum_{k=0}^{\infty}\binom{10+k-1}{k} . (x^3)^k \right ] \\ \\ &\left [ {\color{red}{\bf x^{33}}} \right ] : \begin{cases} (-1)^0.\binom{10}{0}.\binom{10+11-1}{11} \qquad r = 0, k = 11 \\ \\ (-1)^1.\binom{10}{1}.\binom{10+6-1}{6} \qquad r = 1, k = 6 \\ \\ (-1)^2.\binom{10}{2}.\binom{10+1-1}{1} \qquad r = 2, k = 1 \\ \end{cases} \\ \\ &\left [ {\color{red}{\bf x^{33}}} \right ] : \begin{cases} +\binom{10}{0}.\binom{20}{11} \qquad r = 0, k = 11 \\ \\ -\binom{10}{1}.\binom{15}{6} \qquad r = 1, k = 6 \\ \\ +\binom{10}{2}.\binom{10}{1} \qquad r = 2, k = 1 \\ \end{cases} \end{align*}$

 


 

NOTE:

1. $1+x+x^2+x^3+.....x^n = \frac{1-x^{n+1}}{1-x}$

2. $\frac{1}{(1-x)^n} = \sum_{r=0}^{\infty}\binom{n+r-1}{r}.x^r$

3. $\left [ x^{83} \right ]$ means coefficient of $x^{83}$ of the whole expression.

answered by Veteran (51.4k points)  
selected by
+4 votes

Coefficient of x83 in (x5+ x8+ x11+ x14+ x17)10 

= Coefficient of x33 in (1 + x+ x+ x+ x12)   [Just took x5 common so it's x50(1 + x+ x+ x+ x12)10 ]

= Coefficient of x33 in [1-(x3)5 / 1-x ]^10 [Applied sum of GP i.e a(1-rn)/(1-r) where r is common ration and a is first term]

=  Coefficient of x33 in (1 - x15 )10 (1 - x)-10 

= 10C0 * (-1)0 * -10C11 * (-1)11 10C1 * (-1)1 * -10C6 * (-1)610C2 * (-1)2 * -10C1 * (-1)--> now it can be solved :)

To solve futher use extended bionomial theorem ie -nCr  =  (n+r-1)C* (-1)r

answered by Loyal (3.7k points)  
is the answer = 1294080

20C11  - 10* 15C6  + 45* 10C1


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