Let f(x) = 5ˣ/(5ˣ + 5). Then the Sum of the Series f(1/20) + f(2/20) + f(3/20) + ⋯ + f(39/20) Is Equal To

JEE Main 2021 — 25 February, Evening Shift. Previous Year Question.

Problem

Let $f(x)=\dfrac{5^x}{5^x+5}$. Then the sum of the series $f\left(\dfrac1{20}\right)+f\left(\dfrac2{20}\right)+f\left(\dfrac3{20}\right)+\cdots+f\left(\dfrac{39}{20}\right)$ is equal to:

Key insight. The arguments $\frac1{20}$ and $\frac{39}{20}$ add up to 2, and so do every other pair of terms symmetric around the middle — checking whether $f(x)+f(2-x)$ simplifies to something clean is exactly the move this series is built to reward.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Notice that the first and last arguments of the series, $\frac1{20}$ and $\frac{39}{20}$, sum to exactly $2$ — and so does every other matching pair from the two ends working inward. Check whether $f(x)+f(2-x)$ collapses to a constant; if it does, the whole 39-term sum can be found by pairing terms rather than adding them one at a time.

Solution

Step 1 — Test the identity f(x) + f(2−x)

Let $u=5^x$, so $f(x)=\dfrac{u}{u+5}$. Since $5^{2-x}=\dfrac{25}{u}$:

$$f(2-x) = \frac{\frac{25}{u}}{\frac{25}{u}+5} = \frac{25}{25+5u} = \frac{5}{5+u}$$

(dividing numerator and denominator by 5). Adding the two:

$$f(x)+f(2-x) = \frac{u}{u+5}+\frac{5}{u+5} = \frac{u+5}{u+5} = 1$$

So for any $x$, $f(x)+f(2-x)=1$ — this is the identity that makes the whole series collapse.

Step 2 — Pair up the terms of the series

The series runs over $f\left(\frac{k}{20}\right)$ for $k=1,2,\dots,39$. Pair the term for $k$ with the term for $40-k$, since

$$\frac{k}{20}+\frac{40-k}{20} = \frac{40}{20} = 2$$

which is exactly the condition for the identity from Step 1 to apply: $f\left(\frac{k}{20}\right)+f\left(\frac{40-k}{20}\right)=1$.

Step 3 — Identify the pairs and the leftover middle term

Pairing $k=1$ with $k=39$, $k=2$ with $k=38$, and so on up to $k=19$ with $k=21$ gives $19$ pairs, each summing to $1$. The value $k=20$ pairs with $40-20=20$ — itself — so it’s left unpaired:

$$\text{Sum} = \underbrace{19\times1}_{\text{19 pairs}} + f\left(\frac{20}{20}\right) = 19+f(1)$$

Step 4 — Evaluate the leftover term f(1)

$$f(1) = \frac{5^1}{5^1+5} = \frac{5}{10} = \frac12$$

Step 5 — Add it all up

$$\text{Sum} = 19+\frac12 = \frac{38}2+\frac12 = \frac{39}2$$

Answer

Common mistakes

• Trying to substitute all 39 values of $x$ one at a time. This is exactly the trap the series is designed to catch — without spotting the pairing identity, the arithmetic becomes long and error-prone for no reason.
• Miscounting the number of pairs. With $k$ running from 1 to 39, it’s easy to think there are 20 pairs instead of 19 pairs plus one unpaired middle term at $k=20$ — always check which term, if any, pairs with itself.
• Forgetting to add the leftover middle term. After correctly finding 19 pairs each summing to 1, it’s tempting to stop at $19$ rather than adding $f(1)=\frac12$ for the unpaired middle value.

Practise next

• For the same $f(x)=\dfrac{5^x}{5^x+5}$, find $f\left(\dfrac1{10}\right)+f\left(\dfrac2{10}\right)+\cdots+f\left(\dfrac{19}{10}\right)$ using the same pairing idea, and identify the new middle term.