Home
Class 12
MATHS
Prove that (.^(n)C(0))/(1)+(.^(n)C(2))/(...

Prove that `(.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"....."+= (2^(n))/(n+1)`

Text Solution

AI Generated Solution

To prove that \[ \frac{{nC_0}}{1} + \frac{{nC_2}}{3} + \frac{{nC_4}}{5} + \frac{{nC_6}}{7} + \ldots = \frac{{2^n}}{{n+1}}, \] we will follow these steps: ...
Promotional Banner

Topper's Solved these Questions

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.7|9 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.8|10 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.5|8 Videos

  • AREA

    CENGAGE ENGLISH|Exercise Comprehension Type|2 Videos

  • CIRCLE

    CENGAGE ENGLISH|Exercise MATRIX MATCH TYPE|7 Videos

Similar Questions

Explore conceptually related problems

Prove that .^(n)C_(0) + (.^(n)C_(1))/(2) + (.^(n)C_(2))/(3) + "……" +(. ^(n)C_(n))/(n+1) = (2^(n+1)-1)/(n+1) .

Prove that (.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1) .

Prove that (.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1) .

Prove that .^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1) . Hence, prove that .^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N .

if (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a++.^(n)C_(2)a^(2)+ . . .+.^(n)C_(n)a^(n) , then prove that (.^(n)C_(1))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))= Sum of first n natural numbers.

Prove that .^(n)C_(0) - .^(n)C_(1) + .^(n)C_(2) - .^(n)C_(3) + "……" + (-1)^(r) + .^(n)C_(r) + "……" = (-1)^(r ) xx .^(n-1)C_(r ) .

Prove that .^(n-1)C_(3)+.^(n-1)C_(4) gt .^(n)C_(3) if n gt 7 .

If (1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + C_(3) x^(3) + …+ C_(n) x^(n) prove that (C_(0))/(1) + (C_(2))/(3) + (C_(4))/(5) + ...= (2^(n))/(n+1) .

Show that (C_(0))/(1) - (C_(1))/(4) + (C_(2))/(7) - … + (-1)^(n) (C_(n))/(3n +1) = (3^(n) * n!)/(1*4*7…(3n+1)) , where C_(r) stands for ""^(n)C_(r) .

Prove that: (i) r.^(n)C_(r) =(n-r+1).^(n)C_(r-1) (ii) n.^(n-1)C_(r-1) = (n-r+1) .^(n)C_(r-1) (iii) .^(n)C_(r)+ 2.^(n)C_(r-1) +^(n)C_(r-2) =^(n+2)C_(r) (iv) .^(4n)C_(2n): .^(2n)C_(n) = (1.3.5...(4n-1))/({1.3.5..(2n-1)}^(2))