Question

Consider a function $f(k)$ defined for positive integers $k = 1, 2, $; the function satisfies the condition
$f(1) + f(2) + + f(k) = p( p^{k-1} )$ Where $p$ is a fraction i.e. $0<p<1$. Then $f(k)$ is given by:

• A. $p(p)^{k-1}$
• B. $p(1-p)^{k-1}$
• C. $\{ p(1-p) \}^{k-1}$
• D. None of these

Hint:

When a function’s cumulative sum follows a geometric pattern, individual terms form a geometric progression with the same ratio.

Answer 1

The Correct Option is B

We are told that:
$f(1) + f(2) + + f(k) = p( p^{k-1} )$ — Wait, this looks misprinted. A common form for such problems is $f(1) + f(2) + + f(k) = p (1 - p)^{k-1}$ or something involving geometric progressions. Given the multiple-choice options, let us assume the sum up to $k$ is $1 - (1-p)^k$ times some scaling.
If $S(k) = f(1) + + f(k)$ and $S(k) - S(k-1) = f(k)$, then $f(k)$ can be found by subtracting consecutive partial sums. This is a common telescoping approach.
From geometric progression rules, if the total sum is proportional to $(1-p)^k$, then the term $f(k)$ must be proportional to $(1-p)^{k-1}$. Among given choices, the one that matches this pattern is $p(1-p)^{k-1}$.
Indeed, summing $p(1-p)^{n-1}$ from $n=1$ to $k$ gives a GP with ratio $(1-p)$:
Sum = $\frac{p \left[ 1 - (1-p)^k \right]}{1 - (1-p)} = 1 - (1-p)^k$ — a neat form.
Thus, $f(k) = p(1-p)^{k-1}$.