Question

Prove that ${C_0} + 5{C_1} + 8{C_2} + ... + \left( {3n + 2} \right){C_n} = \left( {3n + 4} \right){2^{n - 1}}$?

Answer 1

Hint: In order to prove the above equation , first take the left-hand side of the equation . Rewrite the LHS part in the form of summation and separate it into the terms obtained by the expansion. Now with the help of the binomial expansion result of term ${\left( {x + 1} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {}^n{C_n}$define the value of the 2nd part of the equation by putting $x = 2$. And for the summation of 1st term use the formula of $C(n,r)$.Simplifying all the terms we will get our LHS which is equal to the RHS of the equation.

Complete step by step solution:
 Here in this we have to prove the left-hand Side(LHS) of the equation is equal to the Right-hand side of the equation .
So for this , we are first taking the LHS part of the equation
$LHS = {C_0} + 5{C_1} + 8{C_2} + ... + \left( {3n + 2} \right){C_n}$
Since we have given the nth term of the series. So let’s try to write the series in the summation form
$ = \sum\limits_{r = 0}^n {\left( {3r + 2} \right){}^n{C_r}} $
Now expand the terms using distributive property of multiplication as $\left( {A + B} \right)C = AC + BC$, and separate the summation among the terms. Remember that the constant values can be pull out of the summation, So we get
$ = 3\sum\limits_{r = 0}^n {r{}^n{C_r}} + 2\sum\limits_{r = 0}^n {{}^n{C_r}} $---(1)
Now we know that the binomial expansion of the term ${\left( {x + 1} \right)^n}$is
${\left( {x + 1} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {}^n{C_n}$
Now if we put $x = 1$, we get
${\left( 2 \right)^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .... + {}^n{C_n}$
If we write this thing in the summation form, it becomes
 \[\sum\limits_{r = 0}^n {{}^n{C_r}} = {2^n}\] -----(2)
Putting the above obtained in our original equation(1), we get
$ = 3\sum\limits_{r = 0}^n {r{}^n{C_r}} + 2\left( {{2^n}} \right)$---------(3)
Now recall the formula of
 $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Multiplying both sides with $r$, we get
$
  {r. ^n}{C_r} = \dfrac{{r \times n!}}{{r!(n - r)!}} \\
  {r. ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r)!}} \;
 $
If we try to extract $n$from the $n!$ we will get $n! = n\left( {n - 1} \right)!$
$
  {r. ^n}{C_r} = \dfrac{{n(n - 1)!}}{{(r - 1)!(n - r)!}} \\
  {r. ^n}{C_r} = n\left( {\dfrac{{(n - 1)!}}{{(r - 1)!(n - r)!}}} \right) \;
 $
And we can clearly see that $\dfrac{{(n - 1)!}}{{(r - 1)!(n - r)!}}$is nothing but the ${}^{n - 1}{C_{r - 1}}$, writing this in the above we get
${r. ^n}{C_r} = n\left( {^{n - 1}{C_{r - 1}}} \right)$
Putting the result obtained in above equation in the equation (3)
 \[LHS = 3n\sum\limits_{r = 0}^n {{}^{n - 1}{C_{r - 1}}} + 2\left( {{2^n}} \right)\]
If we try to generalise the result obtained in the equation (2) , we also say that \[\sum\limits_{r = 0}^n {{}^{n - 1}{C_{r - 1}}} = {2^{n - 1}}\]
 \[LHS = 3n\left( {{2^{n - 1}}} \right) + 2\left( {{2^n}} \right)\]
Trying to write ${2^n}$in the form of ${2^{n - 1}}$
 \[LHS = 3n\left( {{2^{n - 1}}} \right) + 4\left( {{2^{n - 1}}} \right)\]
Taking common${2^{n - 1}}$from both the term , our LHS now becomes
 \[LHS = \left( {3n + 4} \right){2^{n - 1}}\]
Now taking the RHS part of the original equation,
  \[RHS = \left( {3n + 4} \right){2^{n - 1}}\]
$LHS = RHS$
Hence, proved

Note: 1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by $n!$.
2.Permutation: Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.
If n and r are positive integers such that $1 \leqslant r \leqslant n$, then the number of all permutations of n distinct or different things, taken r at one time is denoted by the symbol $p(n,r)\,or{\,^n}{P_r}$.
$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
3.Combinations: Each of the different selections made by taking some or all of a number of objects irrespective of their arrangement is called a combination.
The combinations number of n objects, taken r at one time is generally denoted by
$C(n,r)\,or{\,^n}{C_r}$
Thus, $C(n,r)\,or{\,^n}{C_r}$= Number of ways of selecting r objects from n objects.
$C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$