Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The coefficient of x^n-2 in the polynomial (x-1)(x-2)(x-3)..........(x-n) is

Answers (3)

best_answer

Please specify coefficient of $\:\:(1)x^{n-2}\:\:or\:\:(2)x^n-2$

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

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

In general, this is represented by :

$\\(x-a_1)(x-a_2)(x-a_3).......(x-a_n)=x^n+(-1)^1\sum a_1x^{n-1}+(-1)^2\sum a_1a_2x^{n-2}+(-1)^3\sum a_1a_2a_3x^{n-3}+.........(-1)^na_1a_2a_3.......a_n$

Posted by

himanshu.meshram

View full answer

Required sum can be represented as:

$\\Coefficient \:of\:x^{n-2}\:is \:sum\:of\:roots\:taken\:2\:at\:a\:time=Required\:sum\\(1+2+3+4+.......+n)^2=1^2+2^2+3^2+4^2+.......n^2+2(Required\:sum)\\ \because sum\:of\:first\:n\:natural\:no\:\:is\:\:\frac{n(n+1)}{2}\\andsum\:of\:square\;of\;first\;n\:natural\:no\:is\:\frac{n(n+1)(2n+1)}{6}\\Solving\:above\:by\:using\:this\:formula\:you\:will\:get\:Coefficient \:of\:x^{n-2}$ $\\Coefficient \:of\:x^{n-2}\:is \:sum\:of\:roots\:taken\:2\:at\:a\:time=Required\:sum\\(1+2+3+4+.......+n)^2=1^2+2^2+3^2+4^2+.......n^2+2(Required\:sum)\\ \because sum\:of\:first\:n\:natural\:no\:\:is\:\:\frac{n(n+1)}{2}\\andsum\:of\:square\;of\;first\;n\:natural\:no\:is\:\frac{n(n+1)(2n+1)}{6}\\Solving\:above\:by\:using\:this\:formula\:you\:will\:get\:Coefficient \:of\:x^{n-2}$

Posted by

himanshu.meshram

View full answer

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question