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If (1 + x)^(n) = C(0) + C(1) x + C(2) x...

If ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) +…+ C_(n) x^(n)` , find the values of the following .
`sum_(i=0)^(n) sum_(j=0)^(n) (C_(i) + C_(j))`

A

`(n+1)^(2)`

B

`(n+1) 2 ^(n+1)`

C

`n2^(n)`

D

`n2^(n+1)`

Text Solution

Verified by Experts

The correct Answer is:
B

We have,
`sum_(r=0)^(n)sum_(s=0)^(n)(C_(r) +C_(s))`
`sum_(r=0)^(n)sum_(s=0)^(n)C_(r) +sum_(r=0)^(n)sum_(s=0)^(n)C_(s)`
=`sum_(r=0)^(n)(sum_(s=0)^(n)C_(r)) +sum_(r=0)^(n)(sum_(s=0)^(n)C_(s))`
`sum_(r=0)^(n)2^(n)+sum_(s=0)^(n)2^(n)`
= `(n+1) 2^(n) + (n+1)2^(n) = 2 (n+1)2^(n) = (n+1) 2^(n+1)`.
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OBJECTIVE RD SHARMA ENGLISH-BINOMIAL THEOREM AND ITS APPLCIATIONS -Chapter Test
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  3. The expression [x+(x^(3)-1)^((1)/(2))]^(5)+[x-(x^(3)-1)^((1)/(2))]^(...

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  4. The coefficient of x^(53) in the expansion sum(m=0)^(100)^(100)Cm(x-3)...

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  5. If (1 + x)^(n)= C(0) + C(1) x C(2) x^(2) + …+ C(n) x^(n) , prove th...

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  6. Find the numerically grates term in the expansion of 3-5x^(15)w h e nx...

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  7. In the expansion of (1+x)^(50), find the sum of coefficients of odd po...

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  8. Find the position of the term independent of x in the expansion of (sq...

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  9. If the coefficients of x^(7) and x^(8) in the expansion of (2+x/3)^(n)...

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  10. If the rth term in the expansion of (x/3-2/x^(2))^(10 contains x^(4), ...

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  11. If the third in the expansion of [x + x^(logx)]^(6) is 10^(6) , th...

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  12. the value of x , for which the 6th term in the expansions of[2^log2sqr...

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  13. If the coefficients of (p+1)th and (P+3)th terms in the expansion of (...

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  14. about to only mathematics

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  15. The value of C(0)+3C(1)+5C(2)+7C(3)+….+(2n+1)C(n) is equal to :

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  16. Find the following sum : (1)/(n!) + (1)/(2!(n-2)!) + (1)/(4!(n-4)!)+...

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  17. The coefficient of x^(n) y^(n) in the expansion of [(1 + x)(1+y) (x...

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  19. Find the ratio of the coefficient of x^(15) to the term independent of...

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  20. Find the number of terms in the expansion of (x+y+z)^(n).

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  21. In the expansion of (1+x)^30 the sum of the coefficients of odd powers...

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