Home
Class 12
MATHS
Prove that .^(n)C(0) +5 xx .^(n)C(1) + 9...

Prove that `.^(n)C_(0) +5 xx .^(n)C_(1) + 9 xx .^(n)C_(2) + "…." + (4n+1) xx .^(n)C_(n) = (2m+1) 2^(n)`.

Text Solution

AI Generated Solution

To prove that \[ \binom{n}{0} + 5 \cdot \binom{n}{1} + 9 \cdot \binom{n}{2} + \ldots + (4n + 1) \cdot \binom{n}{n} = (2n + 1) \cdot 2^n, \] let's denote the left-hand side as \( S \): ...
Promotional Banner

Topper's Solved these Questions

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.5|8 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.6|10 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.3|7 Videos

  • AREA

    CENGAGE ENGLISH|Exercise Comprehension Type|2 Videos

  • CIRCLE

    CENGAGE ENGLISH|Exercise MATRIX MATCH TYPE|7 Videos

Similar Questions

Explore conceptually related problems

Find the sum .^(n)C_(0) + 2 xx .^(n)C_(1) + xx .^(n)C_(2) + "….." + (n+1) xx .^(n)C_(n) .

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 .

Find the sum .^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "……" + n xx .^(n)C_(n) .

Find the sum 1 xx 2 xx .^(n)C_(1) + 2 xx 3 xx .^(n)C_(2) + "….." + n xx (n+1) xx .^(n)C_(n) .

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)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: (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))

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