recategorized by
634 views
1 votes
1 votes

The following sum of $n+1$ terms $$2 + 3 \times \begin{pmatrix} n \\ 1 \end{pmatrix} + 5 \times \begin{pmatrix} n \\ 2 \end{pmatrix} + 9 \times \begin{pmatrix} n \\ 3 \end{pmatrix} + 17 \times \begin{pmatrix} n \\ 4 \end{pmatrix} + \cdots$$ up to $n+1$ terms is equal to

  1. $3^{n+1}+2^{n+1}$
  2. $3^n \times 2^n$
  3. $3^n + 2^n$
  4. $2 \times 3^n$
recategorized by

1 Answer

4 votes
4 votes

$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\cdots+\binom{n}{n}x^n$
Putting $x=1, 2~$ yields respectively
$2^n=1+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\cdots+\binom{n}{n}  \tag{i}$

and

$3^n=1+\binom{n}{1}\times2+\binom{n}{2}\times2^2+\binom{n}{3}\times2^3+\binom{n}{4}\times2^4+\cdots+\binom{n}{n}\times2^n \tag{ii}$

Adding no$\mathrm{(i)}$ and no$\mathrm{(ii)}$,

$\begin{align} 2^n+3^n&=\scriptsize 2+\binom{n}{1}\times(1+2)+\binom{n}{2}\times(1+2^2)+\binom{n}{3}\times(1+2^3)+\binom{n}{4}\times(1+2^4)+\cdots+\binom{n}{n}\times(1+2^n)\\ &=\scriptsize 2+3\times\binom{n}{1}+5\times\binom{n}{2}+9\times\binom{n}{3}+17\times\binom{n}{4}+\cdots+(2^n+1)\times\binom{n}{n} \end{align}$

Definitely it has $n+1$ terms.

 

So the correct answer is C.

Related questions

1 votes
1 votes
1 answer
1
Arjun asked Sep 23, 2019
627 views
The sum $\sum_{k=1}^n (-1)^k \:\: {}^nC_k \sum_{j=0}^k (-1)^j \: \: {}^kC_j$ is equal to $-1$$0$$1$$2^n$
2 votes
2 votes
2 answers
2
2 votes
2 votes
3 answers
3
Arjun asked Sep 23, 2019
737 views
$^nC_0+2^nC_1+3^nC_2+\cdots+(n+1)^nC_n$ equals$2^n+n2^{n-1}$$2^n-n2^{n-1}$$2^n$none of these
2 votes
2 votes
2 answers
4
Arjun asked Sep 23, 2019
660 views
The sum of the series $\dfrac{1}{1.2} + \dfrac{1}{2.3}+ \cdots + \dfrac{1}{n(n+1)} + \cdots $ is$1$$1/2$$0$non-existent