IIT JEE , AIEEE Forum - Mathematics , Algebra

pds

WHAT WILL BE THE COEFFICIENT OF X ⁿ IN THE EXPANSION OF (1 +X+2X ² +3X ³ +........+n Xⁿ)²

sg_90

is the answer = (n+(n-1)+2*(n-2)+........+(n-1)*1+n)....

chander_singla06

pl explain in detail

sg_90

see it goes like this

(1+x+2x²+3x³+......nxⁿ)(1+x+2x²+3x³+....nxⁿ⁾

here u have to find the coeff. of xⁿ

here u find dat xⁿ coeff..comes as n-1+1,n,n-2+2(here i am talking about addition of powers) there fore my above pist give s the answer

yahiyafirdous

perhaps you given series is wrong, the n-Th term should be nx^n-1

If this is the case then answer is (n+1)

pds

no.the given series is correct and someone please tell the ANSWER.

sg_90

answer in the above post see properly!!!!!

amar.gupta

good work sg_90

given series is

(1+x+2x²+3x³+......nxⁿ)(1+x+2x²+3x³+....nxⁿ⁾

let say A * B

now you have to find the coeff. of xⁿif you closely observe you will find that xⁿ

will appear on expension of above when x⁰i..e. first term of A will be multiplied by term containing xⁿin B.similarlly when x of A multiplied by x^n-1of B and x²multiplied by x^n-2and so on.....

so you have to just write down the corresponding coefficient of the same:

i.e. 1*(n) + 1*(n-1) + 2*(n-2) ................+n*(1)

NOTE:

[coefficient in bracket correspond to B. as 1 is coeff. of x⁰in A and (n) is coeff. of xⁿin B ]

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