Question

The value of \[{{3}^{n}}{{C}_{0}}-{{8}^{n}}{{C}_{2}}+{{13}^{n}}{{C}_{3}}+.......+\text{ n term}\] is
$\begin{align}
  & a)0 \\
 & b){{3}^{n}} \\
 & c){{5}^{n}} \\
 & d)\text{ none of these} \\
\end{align}$

Answer 1

Hint: Now first we will write the general term of the series given and hence convert the series in the form of summation. Now we will open the bracket to form an equation with two terms.
Now we know that according to binomial theorem ${{\left( 1-x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+{{.....}^{n}}{{C}_{n}}{{x}^{n}}$
Substituting x = 1. In the equation we get the value of the first term in the equation. Now again consider the equation ${{\left( 1-x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+{{.....}^{n}}{{C}_{n}}{{x}^{n}}$ . Differentiating the equation with x we will obtain a new equation. Now in this equation we will substitute x = to find the value of the second term. Hence we will add the two terms to find the value of the given series.

Complete step by step answer:
Now consider the given expression\[{{3}^{n}}{{C}_{0}}-{{8}^{n}}{{C}_{2}}+{{13}^{n}}{{C}_{3}}+.......+\text{ n term}\].
Now we can see that the coefficient of the terms in the expression are 3, 8, 13, …
Now we can see that these are in AP with difference 5 and first terms as 3.
Now hence its ${{r}^{th}}$ term will be 3 + 5r
Hence the expression can be written as $\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3+5r \right)}{{.}^{n}}{{C}_{r}}$
Now let us open the bracket,
$\sum\limits_{r=1}^{n}{{{\left( -1 \right)}^{r}}\left( 3+5r \right)}{{.}^{n}}{{C}_{r}}=\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3 \right)}{{.}^{n}}{{C}_{r}}+\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 5r \right)}{{.}^{n}}{{C}_{r}}........................\left( 1 \right)$
Now consider the expression.
$\begin{align}
  & \sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3 \right)}{{.}^{n}}{{C}_{r}}=3\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}}{{.}^{n}}{{C}_{r}} \\
 & \Rightarrow \sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3 \right)}{{.}^{n}}{{C}_{r}}=3\left( ^{n}{{C}_{0}}{{-}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+{{.....}^{n}}{{C}_{n}} \right)......................\left( 2 \right) \\
\end{align}$
Now we know that the binomial expansion of ${{\left( 1-x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+{{.....}^{n}}{{C}_{n}}{{x}^{n}}$
Substituting x = 1 in the above equation we get,
\[0{{=}^{n}}{{C}_{0}}{{-}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+{{.....}^{n}}{{C}_{n}}..........................\left( 3 \right)\]
Now substituting equation (3) in equation (2) we get,
$\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3 \right)}{{.}^{n}}{{C}_{r}}=0..............................\left( 4 \right)$
Now again consider the equation ${{\left( 1-x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+{{.....}^{n}}{{C}_{n}}{{x}^{n}}$
Differentiating the above equation on both sides we get,
$n{{\left( 1-x \right)}^{n-1}}={{-}^{n}}{{C}_{1}}+{{2}^{n}}{{C}_{2}}{{x}^{1}}-{{3}^{n}}{{C}_{3}}{{x}^{2}}+.....{{n}^{n}}{{C}_{n}}{{x}^{n-1}}$
Now again let us substitute x = 1 in the above equation.
Hence we get,
$\begin{align}
  & 0={{-}^{n}}{{C}_{1}}+{{2}^{n}}{{C}_{2}}-{{3}^{n}}{{C}_{3}}+.....{{n}^{n}}{{C}_{n}} \\
 & \Rightarrow 0=\sum\limits_{r=1}^{n}{{{\left( -1 \right)}^{r}}.r{{.}^{n}}{{C}_{r}}}...............................\left( 5 \right) \\
\end{align}$
Now by substituting the values obtained in equation (4) and equation (5) in the equation (2) we get,
$\sum\limits_{r=0}^{n}{{{\left( -1 \right)}^{r}}\left( 3+5r \right)}{{.}^{n}}{{C}_{r}}=0$
Hence the value of \[{{3}^{n}}{{C}_{0}}-{{8}^{n}}{{C}_{2}}+{{13}^{n}}{{C}_{3}}+.......+\text{ n term}\] is equal to 0.

So, the correct answer is “Option a”.

Note: Now note that in the given expression where a = 3 and d = 5 ${{r}^{th}}$ term of an AP is given by $3+\left( r-1 \right)5=5r-2$ . But since we have the numbering starting from 0 and not 1 then to find ${{r}^{th}}$ term we will substitute r + 1 in the formula hence we get $3+5r$ .