Let `1+sum_(r=1)^(10)(3^r.^(10)C_r+r.^(10)C_r)=2^(10)(alpha. 4^5+beta)` where `alpha,beta in N` and `f(x)=x^2-2x-k^2+1.` If `alpha,beta` lies between the roots of `f(x)=0` , then find the smallest positive integral value of `kdot`

Answer

5

Explanation

We have `1+underset(r=1)overset(10)sum(3^(r)..^(10)C_(r)+r..^(10)C_(r))`
`1+underset(r=1)overset(10)sum3^(r)..^(10)C_(r)+10underset(r=1)overset(10)sum.^(9)C_(r-1)`
`=1+4^(10)-1+10xx2^(9)`
`=4^(10)+5xx2=2^(10)(4^(5)+5)`
`= 2^(10)(alpha.4^(5)+beta)`, so `alpha = 1` and `beta = 5`

Now `f(1) lt 0` and `f(5) lt 0`
`f(1) ,t 0 rArr - k^(2) lt 0 rArr k ne 0`
and `f(5) lt 0`
`rArr 16-k^(2) lt 0`
or `k^(2) - 16 gt 0`
`rArr k in (-oo,4)uu(4,oo)`
Hence, the smallest positive integral value of `k = 5`.