IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

the no. of terms in the expansion

(x²+1+1/x²)ⁿ,n

N is:

(A) 2n (B) 3n

(C) 2n+1 (D)3n+1

sir pls give the answer with proper expalnation.

ankurgupta91

(x^2+1+1/x^2)^n
= [(x+1/x)^2 -1 ]^n
nw expanding it we get highest order terms as
(x+1/x)^2n - (x+1/x)^(2n-1) +.........
so the no. of terms in the expansion are 2n+1
ans is (C)
thanku

ashish040191

I have a shortcut to solve this problem

...................................................

......................,,,,,.,.,.,.,.,.,.,.,.,.,.,.

No. of terms in the expansion of

(a+b+c+d)n = ^n+m-1 Cn , Where

m= no. of terms in the bracket

Apply rhis.

Plz rate if u liked the formula.................

iitkgp_bipin

(x² + 1 + 1/x²)ⁿ = [(x + 1/x)² - 1]ⁿ

= (x + 1/x)²ⁿ - (x + 1/x)^2n-1 + (x + 1/x)^2n-2 + ....................

Hence it has 2n+1 terms.

rajat

hey . plzz. tell me y is my method wroong

(x^2 + 1 + 1/x^2)^n

Tr+1 = C(n,r)(x^2 + 1/x^2)^r

now (x^2 + 1/x^2)^r has r+1 terms.

=> Tr+1 has r+1 terms

total terms are by varying r from 0 to n-1

1 + 2 +3.........n =>n(n+1)/2

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