IIT JEE , AIEEE Forum - Mathematics , Algebra

Srijan_theDON

Show that

3.C₀+7.C₁+11.C₂+???.+(4n+3)C_n=(2n+3) 2ⁿ

titun

Hiii,

The (r+1)th term of the given series is (4r+3).nCr = 4r.nCr + 3.nCr

Therefore, the given series can be written as the following :

3. _{[r = 0]}^{[r = n]} nCr + 4. _{[r = 0]}^{[r = n]} r. nCr

= 3. (1+1)^n + 4z ( where z = _{[r = 0]}^{[r = n]} r. nCr )

= 3. 2^n + 4z

In the above step, _{[r = 0]}^{[r = n]} nCr = nC0+nC1+nC2+.......+nCn

This is easily calculated by putting x=1 in the following

(1+x)^n = nC0 + nC1.x + nC2.x^2 +.....................+ nCn.x^n

i.e 2^n = = nC0+nC1+nC2+.......+nCn = _{[r = 0]}^{[r = n]} nCr

Again, (1+x)^n = nC0 + nC1.x + nC2.x^2 +.....................+ nCn.x^n

Differentiating both sides w.r.t x and then putting x = 1, we have,

n.2^(n-1) = z

Therefore, 4z = 2n. 2^n

Hence, the given expression = (2n+3).2^n

This was quite easy. Wasn't it !!