Animated Solution for Mathematics - Sequence and Series: Let ${\sum }_{k=1}^{10}f(a+k)=16(2^{10}-1)$, 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

Imagine you are standing at the edge of a mathematical landscape where functions behave not just as static graphs, but as dynamic, self-replicating entities. The problem provides the functional equation $f(x+y)=f(x)f(y)$.

This is the defining characteristic of exponential growth. It dictates that the function value at a sum of two inputs is the product of the function values at those individual inputs. Given $f(1)=2$, we possess the "DNA" of the function.

If we set $x=1$ and $y=1$, the equation yields $f(1+1)=f(1)\cdot f(1)$, which is $2\cdot 2=2^2$. Extending this to $f(3)$, we find:

The exponent of the base $2$ perfectly matches the input value. By the principle of mathematical induction, we conclude that for any natural number $x$:

We now turn our attention to the summation ${\sum }_{k=1}^{10}f(a+k)$. Substituting our derived function, this becomes ${\sum }_{k=1}^{10}{2}^{a+k}$.

Using the laws of exponents, we know $2^{a+k}=2^a\cdot 2^k$. Since $2^a$ is constant with respect to the index $k$, we factor it out:

The summation ${\sum }_{k=1}^{10}{2}^k$ represents the series $2^1+2^2+2^3+\cdots +2^{10}$. This is a geometric progression where the first term $A=2$, the common ratio $R=2$, and the number of terms $n=10$.

Using the sum formula $S_n=\frac{A(R^n-1)}{R-1}$, we calculate:

We equate our expression to the given value: $2^a\cdot 2(2^{10}-1)=16(2^{10}-1)$. Since $(2^{10}-1)$ is non-zero, we cancel it from both sides:

By comparing the exponents, we find $a+1=4$. Therefore, the final solution is: