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The sum $\sum_{k=1}^n (-1)^k \:\: {}^nC_k \sum_{j=0}^k (-1)^j \: \: {}^kC_j$ is equal to

1. $-1$
2. $0$
3. $1$
4. $2^n$

recategorized | 28 views
0
$A$ ?
0
how ?
0
Just substituted the random values.
0

@`JEET For which value, you are getting (A) ? I am getting (B)

0
may be I committed some mistake.

Will check again.
+2
okay. May be solved by writing inner summation as binomial expression.
+1
Yes!

+1 vote

From the binomial theorem,

$\displaystyle (1+x)^k=\sum_{j=0}^{k} {{}^{k}\textrm{C}_{j}}~x^j$

Putting $x=-1$ to equation above yields,

$\displaystyle (1-1)^k=\sum_{j=0}^{k} {{}^{k}\textrm{C}_{j}}(-1)^j\\ \displaystyle \Rightarrow \sum_{j=0}^{k} (-1)^j~{{}^{k}\textrm{C}_{j}}=0$

Now

$\displaystyle \sum_{k=1}^{n}(-1)^k~{{}^{n}\textrm{C}_{k}} \sum_{j=0}^{k} (-1)^j~{{}^{k}\textrm{C}_{j}}=\sum_{k=1}^{n}(-1)^k~{{}^{n}\textrm{C}_{k}} \times 0=0$

So the correct answer is B.

by Active (3.1k points)

+1 vote