ISI2013-PCB-A-2

Question

Find the value of $ \Sigma ij$, where the summation is over all integers $i$ and $j$ such that $1 \leq i < j \leq 10$.

Answer 1

Given: $1\leq  i < j \leq 10$

$\sum ij = ( 1.2 +1.3 + \dots +1.10 ) + ( 2.3 + 2.4 + \dots +2.10)$
 $+(3.4 +3.5+\dots + 3.10) \dots (8.9+8.10) + 9.10$

$= 1.(2+3+4\dots +10) + 2.(3+4 +\dots +10) + 3.(4+5+ \dots + 10) +4. (5+6+ \dots + 10)$
 $+5. (6+7+\dots + 10) +6. (7+8+\dots+10) +7(8+9+10) + 8(9+10) + (9.10)$

We knows Sum of first Natural Numbers= $\dfrac{n(n+1)}{2}$

now,

$=\Bigg[1.\left[ \dfrac{n(n+1)}{2}-1\right] +2.\left[\dfrac{n(n+1)}{2}-3\right] +3.\left[\dfrac{n(n+1)}{2}-6\right]$
$+4.\left[\dfrac{n(n+1)}{2}-10\right]+5.\left[\dfrac{n(n+1)}{2}-15\right] +6.\left[\dfrac{n(n+1)}{2}-21\right]$
$ +7.\left[\dfrac{n(n+1)}{2}-28\right] + 8.\left[\dfrac{n(n+1)}{2}-36\right] +9.\left[\dfrac{n(n+1)}{2}-45\right]\Bigg]$

$= 55 . \left[1 + 2 + \dots 9\right] - \sum_{i=1}^{9} i. i. \dfrac{(i+1)}{2}$
$ = 55. 45 - \dfrac{1}{2} \sum_{i=1}^{10} i^3 + i^2$
$= 2475 - \dfrac{1}{2} \left(45^2 + 9.10.\dfrac{19}{6}\right)$
$ = 1320.$

Comments

I added at end of answer..

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yes,Thank You..Now it's look good.

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Hi @LeenSharma or @Arjun ji,

Please correct summation limit in third last line it should be from i=1 to i=9.

Answer 2

=1(2+3+4+5+6+....+10)+2(2+3+..........10)+..............+9(2+3+4+5+6+....+10)

=(1+2+3+4+......+9)(2+3+......+10)  {sum of digits from 2 to 10 is (n(n+1)/2)-1}

=45*54

=2430