Home
Class 12
MATHS
Find the following sums : (i) .^(n)C(0...

Find the following sums :
(i) `.^(n)C_(0)-.^(n)C_(2)+.^(n)C_(4)-.^(n)C_(6)+"....."`
(ii) `.^(n)C_(1)-.^(n)C_(3)+.^(n)C_(5)-.^(n)C_(7)+"...."`
(iii) `.^(n)C_(0)+.^(n)C_(4)+.^(n)C_(8)+.^(n)C_(12)+"....."`
(iv) `.^(n)C_(2) + .^(n)C_(6) + .^(n)C_(10)+.^(n)C_(14)+"......"`
(v) `.^(n)C_(1) + .^(n)C_(5)+.^(n)C_(9)+.^(n)C_(13)+"...."`
(vi) `.^(n)C_(3) + .^(n)C_(7) + .^(n)C_(11) + .^(n)C_(15) + "....."`

Text Solution

AI Generated Solution

To solve the given sums using the Binomial Theorem, we will follow the steps outlined in the video transcript. We will consider the binomial expansion of \( (1 + x)^n \) and substitute appropriate values to derive the required sums. ### Step-by-Step Solution: 1. **Binomial Expansion**: The binomial expansion of \( (1 + x)^n \) is given by: \[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k ...
Promotional Banner

Topper's Solved these Questions

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Example|10 Videos
  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Concept Application Exercise 8.1|17 Videos
  • AREA

    CENGAGE ENGLISH|Exercise Comprehension Type|2 Videos
  • CIRCLE

    CENGAGE ENGLISH|Exercise MATRIX MATCH TYPE|7 Videos

Similar Questions

Explore conceptually related problems

If .^(n)C_(8)=.^(n)C_(6) , then find .^(n)C_(2) .

Evaluate .^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"...."+(n-1).^(n)C_(n-2).^(n)C_(n) .

If .^(n)C_(9)=.^(n)C_(7) , find n.

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

If .^(n)C_(5) = .^(n)C_(7) , then find .^(n)P_(3)

If .^(n)C_(8) = .^(n)C_(2) , find .^(n)C_(2) .

Find .^(n)C_(1)-(1)/(2).^(n)C_(2)+(1)/(3).^(n)C_(3)- . . . +(-1)^(n-1)(1)/(n).^(n)C_(n)

Find the value of n: (i) .^(n)C_(10)=^(n)C_(16) (ii) .^(15)C_(n) =^(15)C_(n+3) (iii) .^(10)C_(n) =^(10)C_(n+2) (iv) .^(25)C_(3n) =.^(25)C_(n+1) (v) .^(n)C_(r) =.^(n)C_(r-2)

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 ) .

The A.M. of the series .^(n)C_(0), .^(n)C_(1), .^(n)C_(2),….,.^(n)C_(n) is