Latest TCS Aptitude Question SOLUTION: Find sum of (1^2)*28C1+(2^2)*28C2+.........upto(28^2)28C28.

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Probability

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40028825) TCS Numerical Ability question Keep an EYE

Find sum of (1^2)*28C1+(2^2)*28C2+.........upto(28^2)28C28.

Asked In TCS Ankit Bansal (10 years ago)

Unsolved Read Solution (2)

Is this Puzzle helpful?   (1)   (6) Submit Your Solution Probability

Read Solution (2) :

Subhajit Dutta  (10 years ago)

ans is 406.2^27
We know,(1+x)^n=C0+C1x+C2x^2+.....+Cnx^n .......(1)
Differentiating w.r.t x we get
n(1+x)^(n−1)=C1+2C2x+3C3x^2+.....+nCnx^(n−1)
Multiplying by x on both sides,
x.n(1+x)^(n−1)=x.C1+2C2x^2+3C3x^3+.....
Now again differentiating w.r.t to x
n(1+x)^(n−1)+n(n−1)x(1+x)^(n−2)=C0+2^2C1x+3^2C2x^2+4^2C3x^3+.....
Putting x = 1, we get
n(n+1)2^(n−2)=C1+2^2C2+3^2C3+4^2C4+.....
Now substituting n=28
28(28+1)2^(28−2) = 812.2^26 = 406.2^27

Is this solution Helpfull? Yes (31) | No (1)  

samarth shrivastava  (10 years ago)

sum=n/2(a+l)
a:first term
l:last term
sum=11368

Is this solution Helpfull? Yes | No (3)  

Please Login to Read Solution.