search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
1 vote
136 views

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. $2 \times 3^n$
in Combinatory
recategorized by
136 views
0
C?

1 Answer

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.

2
good answer,
2
I was stuck on this.

Thanks.
4
Let n=1 take two term sum=2+3=5

In and only C option satisfied
1

@amit166
Yeah. That's the trick in the exam hall. πŸ‘

Related questions

1 vote
1 answer
1
94 views
The sum $\sum_{k=1}^n (-1)^k \:\: {}^nC_k \sum_{j=0}^k (-1)^j \: \: {}^kC_j$ is equal to $-1$ $0$ $1$ $2^n$
asked Sep 23, 2019 in Combinatory Arjun 94 views
2 votes
2 answers
2
345 views
Let $(1+x)^n = C_0+C_1x+C_2x^2+ \dots + C_nx^n$, $n$ being a positive integer. The value of $\left( 1+\dfrac{C_0}{C_1} \right) \left( 1+\dfrac{C_1}{C_2} \right) \cdots \left( 1+\dfrac{C_{n-1}}{C_n} \right)$ is $\left( \frac{n+1}{n+2} \right) ^n$ $ \frac{n^n}{n!} $ $\left( \frac{n}{n+1} \right) ^n$ $ \frac{(n+1)^n}{n!} $
asked Sep 23, 2019 in Combinatory Arjun 345 views
2 votes
3 answers
3
208 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
asked Sep 23, 2019 in Combinatory Arjun 208 views
1 vote
2 answers
4
200 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
asked Sep 23, 2019 in Numerical Ability Arjun 200 views
...