Question

find the value:
1^2.C1+2^2.C2+3^2.C3+.....+n^2.Cn find the value:
1^2.C1+2^2.C2+3^2.C3+.....+n^2.Cn @Mathematics

Answer 1

ans is 1+6+84+1820+..............................

Answer 2

so?? what will be the answer in term of n??

Answer 3

and how could u get 6, 84 and so on??

Answer 4

C1, C2, C3 are short forms of nC1, nC2, nC3 and so on..

Answer 5

every time u are having 2 before your C so anything such as \[^{2}C_{3}\]
it is meaningless

Answer 6

u find in which series the eqn is i.e,1+6+84+................... then u will get answer in terms of n

Answer 7

2C3, 2C4,..........and so on is meaning less

Answer 8

all i wanted to mean: 1^2 * nC1

Answer 9

where n is 1,2,3,4..............and so on right

Answer 10

no, n is a number. u have to find the result in terms of n
(1^2)*nc1+(2^2)*nc2+....

Answer 11

find the result upto n terms

Answer 12

ok so u having two series right .........

Answer 13

hmm, kinda

Answer 14

\[\sum_{k=1}^{n}k ^{2}*nCk\]

Answer 15

i guess, it is more clear

Answer 16

but what about n?

Answer 17

n is a fixed unknown number, u have to find the answer in term of n

Answer 18

the lower limit of k is 1 and the upper limit is n