Latest TCS Aptitude Question SOLUTION: What is $\sum\limits_{K = 0}^{28} {{K^2}(_K^{28}C)} $ where $_K^{28}C$ is the number of ways of choosing k items from 28 items? a. 406 * ${2^{27}}$ b. 306 * ${2^{26}}$

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What is $sumlimits_{K = 0}^{28} {{K^2}(_K^{28}C)} $ where $_K^{28}C$ 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 Heena Jain (10 years ago)

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satish  (9 years ago)

Consider (1+x)n=C0+C1x+C2x2+.....+Cnxn .......(1)
Differentiating w.r.t x we get
n(1+x)n−1=C1+2C2x+3C3x2+.....+nCnxn−1
Multiplying by x on both sides,
x.n(1+x)n−1=x.C1+2C2x2+3C3x3+.....
Now again differentiating w.r.t to x
n(1+x)n−1+n(n−1)x(1+x)n−2=C0+22C1x+32C2x2+42C3x3.....
Putting x = 1, we get
n(n+1)2n−2=C1+22C2+32C3+42C4
Now substituting n = 28
28(28+1)228−2 = 812.226 = 406.227

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