ISI2014-DCG-34

Question

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

• A. $3^{n+1}+2^{n+1}$
• B. $3^n \times 2^n$
• C. $3^n + 2^n$
• D. $2 \times 3^n$

Comments

C?

Answer 1

$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\cdots+\binom{n}{n}x^n$
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.

Comments

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

In and only C option satisfied

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@amit166
Yeah. That's the trick in the exam hall. 👍

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@Sachin Mittal 1 @Deepak Poonia

Hi sir, first I solved this question in a simple way by taking n = 3. But it is not a good approach to analyze the question properly.

So sir, as mentioned in the answer, is it the only way to think of this question or is there any other way to think about the answer to this type of question? (Another similar type of question: https://gateoverflow.in/321937/isi2014-dcg-18)