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Let X = (.^(10)C(1))^(2) + 2(.^(10)C(2))...

Let `X = (.^(10)C_(1))^(2) + 2(.^(10)C_(2))^(2) + 3(.^(10)C_(3))^(2) + "……" + 10(.^(10)C_(10))^(2)`, where `.^(10)C_(r ), r in {1,2,"….",10}` denotes binomial coefficients then the value of `(1)/(1430) X` is `"_____"`.

Text Solution

Verified by Experts

We have,
`X=(""^(10)C_(1))^(2)+2(""^(10)C_(2))^(2)+3(""^(10)C_(3))^(2)+...+10(""^(10)C_(10))^(2)`
`impliesX=underset(r=1)overset(10)sumr(""^(10)C_(r))^(2)impliesX=underset(r=1)overset(10)sumr^(10)C_(r)""^(10)C_(r)`
`impliesX=underset(r=1)overset(10)sumrxx(10)/(r)""^(9)C_(r-1)""^(10)C_(r)" "[because""^(n)C_(r)=(n)/(r)""^(n-1)C_(r-1)]`
`impliesX=10underset(r=1)overset(10)sum""^(9)C_(r-1)""^(10)C_(r)`
`impliesX=10underset(r=1)overset(10)sum""^(9)C_(r-1)""^(10)C_(r)" "[because""^(n)C_(r)=""^(n)C_(n-r)]`
`impliesX=10xx""^(19)C_(9)" "[because""^(n-1)C_(r-1)""^(n)C_(n-r)=""^(2n-1)C_(n-1)]`
Now, `(1)/(1430)X=(10xx""^(19)C_(9))/(1430)=(""^(19)C_(9))/(148)=(""^(19)C_(9))/(11xx13)`
`=(19xx17xx16)/(8)=19xx34=646`
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