# ISI2014-DCG-34

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. $2 \times 3^n$

recategorized
0
C?

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

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