IIT JEE , AIEEE Forum - Mathematics , Algebra

gabriel

If (x+1)(x+2)(x+3)....(x+20)=ax²⁰ +bx¹⁹ +cx¹⁸ +....,then

a+b+c=

a)18714

b)20116

c)20826

d)21616

kusum

hi ,

if we use multinomial theorem here we know
a=1
b=1+2+...............+20
c=(sigma)j(sigma)i (i.j)

vish0001

this cane be done like this---------i think so----------

the roots of the equation are -1 , -2.......................-20
thus, u can now find the sum and products of the equation
ya,... i m sure now... procedd like this-- u will get it !
thank u !

aman23iit

hi friend this is a fantastic Q

here is the solution

see you know that the coefficent of x^20 is 1

and if you will closely watch then you will see that coefficent of x^19=1+2+3+.........20

and of x^18 =2+11+33........ which is in 2 degree ap and the assume the n^th term to be a'n³+b'n²+c'n+d' now put the value n=1,2,3,4 and find the value of a',b',c',d' and then apply

a'n³+

b'n²+

c'n+

d'=c a

and hence find the value of a+b+c

bye

plzz rate me

vish0001

aman , is my method correct ?? please cross check !!!!

iitkgp_bipin

Clearly a = 1

b = 1 + 2 + 3 + ...... + 20 = 210

c can be found by taking products of roots taken two at a time.
c = (1)(2+3+......+20) + (2)(3+4+5+....+20) + ....... + (19)(20)

Hence you can found a+b+c

Vish you are correct in your approach.

gabriel

could u explain me how x^19=1+2+3+....+20

vish0001

thank u sir
gabriel this is found out by the theory of equations ! u studied it ??
do tell me ! it comes in the chapter """ theory of equations"""

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