Question

find in ascending powers of x the first three terms of (2+x)(1+x)^8​

Answer 1

Answer:

3

Step-by-step explanation:

S=(x1/2+2x1/41)n

    =∑nCr.(x1/2)n−r.(2x1/41)r

    =∑nCr.(21)r.x(n−r)/2.x−r/4

⇒S=∑r=0nnCr(21)rx(2n−3r)/4

We have to expand in decreasing powers of x.

∴S=nC0(21)0xn/2+nC1(21)1x(2

∴S=nC0(21)0xn/2+nC1(21)1x(2n−3)/4+nC2(21)2x(2n−6)/4+......

First three terms form an A.P.

=nC0,nC1(21)&nC2(21)2 from an A.P.

⇒nC0+nC2(21)2=2(nC1).2

⇒1+2n(n−1)×41=n

⇒(n−1)(n−8)=0

n=1 (∵ there are at least 3 term)

⇒n=8

∴S=∑r=088Cr(21)2x4−3r/4

⇒4−43r=K;      Where K∈I

Now of such values of r ranging from 0 to 8 are :→0,4 and 8

∴ There will be three such terms.

Hence, the answer is 3.