AP ICET 2015 Paper

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Question : 131 of 200

Marks: +1, -0

If

{C}_{{r}}=^{10}{C}_{{r}}

for

0≤r≤10

then

2C_{0}+3C_{1}+4C_{2}+. . ..+11C_{10}=

$2^10$
6656
10240
5632
None of above

Solution:

Given :

C_r=10C_r

for 0 < r < 10

2.C_0+3.C_1+4.C_2+.....+12.C_10

nth term is

(n+2).C_n=n.C_n+2.C_n

∴∑↙{n=0}↖{10}(n+2)⋅C_{n}=∑↙{n=0}↖{10}n⋅C_{n}+∑↙{n=0}↖{10}2.C_{n}

=10.2^{10-1}+2.2^{10}

[

∑↙{n=0}↖{n}n.C_{n}=n.2^{n-1}

and

∑↙{n=0}^{n}C_{n}=2^{n}

]

=10.2^{9}+2^{2}⋅2^{9}

=2^{9}(10+4)=512×14=7168

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