Find the 20th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n$ terms.

Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Miscellaneous Exercise · Q12

2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n

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Concept understanding — Sum Of Products

Sum of Products — The Intuition First

Imagine you're buying a fruit basket. The shop has a rule: you must pick exactly one fruit from each of several groups. Group A has apples and bananas. Group B has oranges and mangoes. How many different baskets can you make?

You'd list them: (apple, orange), (apple, mango), (banana, orange), (banana, mango). That's $2 \times 2 = 4$ baskets.

Now, suppose each basket has a price that depends on which fruits you pick. The total money you'd spend if you bought every possible basket exactly once is the sum of the prices of all those baskets.

That's the core idea of Sum of Products: when you have a situation where you choose one item from each of several independent categories, the total "value" across all possible combinations is the sum of the products of the individual values.

The Precise Statement

\sum_{i_1=1}^{n_1} \sum_{i_2=1}^{n_2} \cdots \sum_{i_k=1}^{n_k} \left( a_{i_1}^{(1)} \cdot a_{i_2}^{(2)} \cdots a_{i_k}^{(k)} \right) = \left( \sum_{i_1=1}^{n_1} a_{i_1}^{(1)} \right) \left( \sum_{i_2=1}^{n_2} a_{i_2}^{(2)} \right) \cdots \left( \sum_{i_k=1}^{n_k} a_{i_k}^{(k)} \right)

In plain English: The sum over all combinations of products equals the product of the sums.

Let that sink in. It's not obvious — it's a beautiful distributive property that works because multiplication distributes over addition.

Why It Works — A Simple Example

Take two small groups:

Group 1: numbers $a_1, a_2$
Group 2: numbers $b_1, b_2$

All possible products: $a_1b_1 + a_1b_2 + a_2b_1 + a_2b_2$

Factor it: $a_1(b_1 + b_2) + a_2(b_1 + b_2) = (a_1 + a_2)(b_1 + b_2)$

That's it. The sum of all products is just the product of the sums. This extends to any number of groups.

Note

This is not the same as "product of sums" (which is a different expression). The order matters: Sum of Products = Product of Sums, but Product of Sums ≠ Sum of Products in general.

Where You'll Meet It

Probability: If you roll two dice, the sum of probabilities of all outcomes is $(\frac16 + \frac16 + \cdots)(\frac16 + \cdots) = 1 \times 1 = 1$ .

Combinatorics: Counting total number of combinations — each group has $n_i$ choices, so total combinations = $n_1 \times n_2 \times \cdots \times n_k$ . That's a special case where each "value" is just 1.

Algebra: Expanding $(x+2)(x+3)$ gives $x^2 + 5x + 6$ — that's a sum of products (each term is a product of one term from each bracket).

Tip

When you see a problem that says "find the sum of all possible products formed by taking one element from each set", immediately think: product of the sums. It saves enormous calculation.

A Common Mistake …

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Each term of the series is the product of two consecutive even numbers, $T_k = (2k)(2k+2) = 4k(k+1)$ ; substituting $k=20$ gives the 20th term as $1680$ .

The series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots$ , continued in the same pattern (" $+\, n$ terms" here just signals the series keeps extending indefinitely in this pattern, the same way NCERT writes an open-ended series elsewhere in this chapter — it is not asking for a formula in $n$ ). The actual question only asks for one specific term: the 20th.

Step 1: Find the general ( $k$ th) term.

Looking at the given terms, the first factor in each product is an even number ( $2, 4, 6, \ldots$ ) and the second factor is the next even number after it ( $4, 6, 8, \ldots$ ). Writing the $k$ th even number as $2k$ , the next even number is $2k+2$ , so:

T_k = (2k)(2k+2) = 4k^2 + 4k = 4k(k+1)

Step 2: Check the formula against the given terms. …

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Q.Find the 20th term of the series 2×4+4×6+6×8+…+n2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n2×4+4×6+6×8+…+n terms.

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Q.Find the 20th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n$ terms.

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