## Basic Algebra for IITJEE Main

Main Problem: For an integer $n \geq 5$, let $S_{n}$ denote the sum of the products of the integers from 1 to n taken three at a time. Then, what is the value of $S_{10}$?

First Hint: Consider the terms in the expansion of $(1+2+3+ \ldots +10)^{3}$.

Second Hint: Classify these terms according to the number of times they occur.

Solution: There are totally 1000 terms in the expansion of

$(1+2+3+ \ldots +10)^{3}$. Terms of the form $ijk$ with i,j,k all distinct appear 6 times each. We are interested in the sum of these terms each such term considered only once. Then, there are terms of the form $i^{2}jk$ with

$i \neq j$. Each such product appears three times. Finally, there are terms of  the form $i^{3}$. Each such term appears once only. Hence,

$A=6S_{10}+3B+C$ equation I

where $A=(\sum_{i=1}^{10})^{3}$ and $C=\sum_{i=1}^{i=10}i^{3}$ and

$\sum_{i \neq j}^{10}i^{2}j$.

Using the well-known formulae given below:

$1+2+3+ \ldots +n = \frac {n(n+1)}{2}$ equation II

$1^{2}+2^{2}+3^{2}+ \ldots + n^{3}= \frac {n^{2}(n+1)^{2}}{4}$ equation III

A and C come out to be respectively $(55)^{3}$ and $55^{2}$.

To calculate B, note that if we add to B, products of the form $i^{2}i$, then we get all the terms of the product

$(1^{2}+2^{2}+3^{2}+ \ldots + n^{2})(1+2+3+\ldots +n)$.

The first factor equals $55 \times 7$ using the formula

$\sum n^{2}=\frac {n(n+1)(2n+1)}{6}$ equation IV

Hence, $B=6 \times (55)^{2}$. Putting these values in equation I, we get

$S_{10}=(1/6)(A-3B-C)=18150$.

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Nalin Pithwa

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