Question

Let $\sum\limits_{k=1}^{10}{f(a+k)=16({{2}^{10}}-1)}$ , where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x, y and f(1)=2, then the natural number ‘a’ is
(a)4
(b)3
(c)16
(d)2

Answer 1

To solve this question, we will first find out the values of f(1), f(2), …. f(10). Then, we will expand the given series in question and will substitute the values of f(1), f(2), …. f(10) in series. And then using summation of G.P we will get the value a.

Complete step by step answer:
Now, in question it is given that f( 1 ) = 2……..( i ) and f( x + y ) = f( x )f( y )
So, at x = 1 and y = 1, we get f( 1 + 1 ) = f ( 2 ),
Or, f ( 2 ) = f( 1 )f( 1 )
and, we know that f( 1 ) =2,
then, $f(2)=2\times 2$
So, f ( 2 ) = 4………( ii )
at x = 2 and y = 1, we get f( 2 + 1 ) = f ( 3 ),
Or, f ( 3 ) = f( 2 )f( 1 )
and, we know that f( 2 ) = 4,
then, $f(3)=4\times 2$
So, f ( 3 ) = 8………( ii )
So, we got f( 1 ) =2, , $f\left( 2 \right)=4={{2}^{2}}$ , $f\left( 3 \right)=8={{2}^{3}}$, ……
So, we can say that $f\left( 10 \right)={{2}^{10}}$
Now, we have $\sum\limits_{k=1}^{10}{f(a+k)=16({{2}^{10}}-1)}$
So, on expanding summation, we get
$f(a+1)+f(a+2)+......+f(a+10)=16({{2}^{10}}-1)$
We can write, f( a + 1 ) = f ( a ) f ( 1 ) and others terms too.
So, we get $f(a)f(1)+f(a)f(2)+......+f(a)f(10)=16({{2}^{10}}-1)$
Taking, f( a ) common, we get
$f(a)\\{f(1)+f(2)+......+f(10)\\}=16({{2}^{10}}-1)$
Putting, values of f ( 1 ), f ( 2 ), ………, f ( 10 ), we get
$f(a)\\{2+{{2}^{2}}+......+{{2}^{10}}\\}=16({{2}^{10}}-1)$
We can see that, $2+{{2}^{2}}+......+{{2}^{10}}$is G.P with first term 2 and common ratio 2 and also we know that sum of n terms of G.P is $\dfrac{a({{a}^{n}}-1)}{a-1}$ and $f(a)={{2}^{a}}$ .
So, summation of $2+{{2}^{2}}+......+{{2}^{10}}=\dfrac{2({{2}^{10}}-1)}{2-1}$
So, we get ${{2}^{a}}\left( \dfrac{2({{2}^{10}}-1)}{2-1} \right)=16({{2}^{10}}-1)$
On solving, we get
${{2}^{a}}\left( 2({{2}^{10}}-1) \right)=16({{2}^{10}}-1)$
On simplifying, we get
${{2}^{a+1}}=16$
Now, we know that ${{2}^{4}}=16$ ,
So, ${{2}^{a+1}}={{2}^{4}}$
On comparing we get
a + 1 = 4
a = 3

So, the correct answer is “Option B”.

Note: To solve such questions, one must know the series formation and what is the sum of difference series such as G.P and A.P. Also, one must read the question twice because there are always some hints in question to do solutions. Calculation mistakes must be avoided.