Latest TCS Aptitude Question SOLUTION: What is ∑ K^2(28cK) where K=0 to 28 and 28cK is the number of ways of choosing k items from 28 items? a. 406 * 2^27 b. 306 * 2^26 c. 28 * 2^27 d. 56 * 2^27

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What is ∑ K^2(28cK) where K=0 to 28 and 28cK is the number of ways of choosing k items from 28 items?

a. 406 * 2^27
b. 306 * 2^26
c. 28 * 2^27
d. 56 * 2^27

Asked In TCS vikas (10 years ago)

Solved RAKESH Read Solution (4)

Is this Puzzle helpful?   (41)   (17) Submit Your Solution Algebra

Read Solution (4) :

RAKESH  (10 years ago)

we know,
(1+x)^n=nc0+nc1.x^1+nc2.x^2+nc3.x^3+...+ncn.x^n

differentiate both sides w.r.t x
n(1+x)^n-1=nc1+2.nc2.x+3.nc3.x^2+...+n.ncn.x^n-1

now multiply both sides by x
x.n(1+x)^n-1=nc1.x+2.nc2.x^2+3.nc3.x^3+...+n.ncn.x^n

again, differentiate both sides w.r.t x
x.n(n-1)(1+x)^n-2 + n(1+x)^n-1 = 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1

n(1+x)^n-2[x(n-1)+(1+x)]= 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1

=> n(1+nx)(1+x)^n-2 = 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1

putting x=1 we get,
=> n(1+n)*2^n-2 = 1^2*nc1+2^2*nc2+3^2*nc3+...+n^2*ncn

so, ∑ K^2(kCk)[K=0 to K=K] = k(k+1)*2^k-2

put k=28 in the above result,

we get, ∑ K^2(28cK) where K=0 to 28 = 28*29*2^26 = 14*29*2^27 = 406*2^27

ans: 406 * 2^27

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aaveri sarkar  (10 years ago)

= n(n+1)/2
=28*29/2
=406
ans is-> a>406*2^27

Is this solution Helpfull? Yes (7) | No (6)  

kulwant singh  (10 years ago)

formula = n(n+1)2^n-2 this is formula i dont know its explaination

Is this solution Helpfull? Yes (5) | No (5)  

aditya  (10 years ago)

guyz plz find more puzzels on this web link :http://adityasrmca.blogspot.in/

Is this solution Helpfull? Yes (1) | No (7)  

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