Get Answers to all your Questions

header-bg qa

Co- efficient  of x ^{15}\: \: in \: \: \left ( 1+ x+ x^3 + x^4 \right )^n 

Option: 1

\sum_{r=0 }^{5} ^nC _{15-3r } {^nC_r}


Option: 2

\sum_{r=0 }^{5} ^n C _{5r}


Option: 3

\sum_{r=0 }^{5} ^n C _{3r}


Option: 4

\sum_{r=0 }^{3} ^n C _{3 -r } ^n C_{5r}


Answers (1)

best_answer

 

Coefficient of x^{R} -

We write general term T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}

and a= f\left ( x \right )

\dpi{120} T_{r+1}=^{n} c_{r}\cdot x^{n-r}\left ( f\left ( x \right ) \right )^{r}

We arrange all of x together and make x^{\left ( n,\,r \right )}

compare : \dpi{120} x^{\left ( n,r \right )}= x^{R}

find r

- wherein

Take a in terms of x.

r can't be negative or fraction.

 

 

co- efficient of    x ^{15 } \: \: in \: \: (1+x+x^3 + x^4 ) ^n \\\\

co-efficient of 

x ^{15 } \: \: in \: \:( 1+ x^3 )^n ( 1+ x ) ^n \\\\ = ^ n C _0 ^ n C _{15}+ ^ n C _1^ n C _{12} + ^ n C _2 ^ n C _9 + ^ n C _3 ^ n C _6 + ^ n C _4 ^ n C _3 + ^ n C _5 ^ n C _0

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE