Question

The evaluated value of ${\sum\nolimits_{i = 0}^n {\sum\nolimits_{j = 1}^n {^n{C_j}} } ^j}{C_i},i \le j$
A. ${3^n} + 1$
B. ${3^n} - 1$
C. ${3^{n + 1}} + 1$
D. None of these

Answer 1

Hint:Here in this question we will apply binomial expansion and combination formula which is mentioned below:-
$\sum\limits_{r = 0}^n {^n{C_r}.{X^r} = [{C_0} + {C_1}X + {C_2}{X^2} + ...{C_n}{X^n}]} $
Combination: -Number of combination of ‘n’ things has taken ‘r’ at a time then combination formula is given by: -
$C(n,r) = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step-by-step answer:
Now we will apply binomial expansion in ${\sum\nolimits_{i = 0}^n {\sum\nolimits_{j = 1}^n {^n{C_j}} } ^j}{C_i},i \le j$
${ \Rightarrow ^n}{C_1}{(^1}{C_0}{ + ^1}{C_1}){ + ^n}{C_2}{(^2}{C_0}{ + ^2}{C_1}{ + ^2}{C_2}) + {....^n}{C_n}{(^n}{C_1}{ + ^n}{C_2}{ + ^n}{C_n})$
Now we will apply combination formula $C(n,r) = \dfrac{{n!}}{{r!(n - r)!}}$
$ \Rightarrow C(1,0) = \dfrac{{1!}}{{0!(1 - 0)!}} = 1$ and $C(1,1) = \dfrac{{1!}}{{1!(1 - 1)!}} = 1$
${ \Rightarrow ^n}{C_1}(1 + 1){ + ^n}{C_2}{(^2}{C_0}{ + ^2}{C_1}{ + ^2}{C_2}) + {....^n}{C_n}{(^n}{C_1}{ + ^n}{C_2}{ + ^n}{C_n})$
Now we will add the inner terms
${ \Rightarrow ^n}{C_1}({2^1}){ + ^n}{C_2}{(^2}{C_0}{ + ^2}{C_1}{ + ^2}{C_2}) + {....^n}{C_n}{(^n}{C_1}{ + ^n}{C_2}{ + ^n}{C_n})$
Similarly we can write other terms in form of 2 with degree 2 and n as it is the same as the previous term expansion.
${ \Rightarrow ^n}{C_1}({2^1}){ + ^n}{C_2}({2^2}) + {....^n}{C_n}({2^n})$
Now we will add and subtract $^n{C_0}({2^0})$ in the above equation.
${ \Rightarrow ^n}{C_0}({2^0}){ + ^n}{C_1}({2^1}){ + ^n}{C_2}({2^2}) + {....^n}{C_n}({2^n}){ - ^n}{C_0}({2^0})$
As we can see that expansion in the form of ${(1 + x)^n} = \sum\limits_{r = 0}^n {^n{C_r}.{X^r} = [{C_0} + {C_1}X + {C_2}{X^2} + ...{C_n}{X^n}]} $ has become. And in place of x there is 2.
$ \Rightarrow {(1 + 2)^n}{ - ^n}{C_0}({2^0})$
Now we will apply combination formula again $C(n,0) = \dfrac{{n!}}{{0!(n - 0)!}} = 1$ and also any term power 0 will be one so ${(2)^0} = 1$
$ \Rightarrow {(3)^n} - 1$

So, the correct answer is “Option B”.

Note:Students may likely make mistakes in doing factorial of zero and can write directly zero without thinking but factorial of zero is one. Also while expanding binomial expansion be careful and remember that expansion very cautiously as without its knowledge these types of questions cannot be done.