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Let 1+ sum(r=1)^(10) (3^(r).""^(10)C(r)+...

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) - k^(2) + 1`. If `alpha, beta` lies between the roots of `f(x) = `, the smalles positive integral value of k is `"____"`.

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The correct Answer is:
5

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`.
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