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.

Please do send your comments, suggestions, etc. 

More later

Nalin Pithwa

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