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Algebra Level 4

∑ c = 1 10 ∑ b = 1 c ∑ a = 1 b a 2 = ? \Large \displaystyle \sum^{10}_{c=1} \sum^{c}_{b=1} \sum^{b}_{a=1}a^{2}= \ ? \


The answer is 3289.

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Akshat Sharda by Akshat Sharda , 21, India

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3 solutions

Let the sum of the expression be S S . Then we have:

S = ∑ c = 1 10 ∑ b = 1 c ∑ a = 1 b a 2 [ ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 ] = ∑ c = 1 10 ∑ b = 1 c b ( b + 1 ) ( 2 b + 1 ) 6 = ∑ c = 1 10 ∑ b = 1 c 2 b 3 + 3 b 2 + b 6 [ ∑ k = 1 n k 3 = ( n ( n + 1 ) 2 ) 2 ] = ∑ c = 1 10 1 6 ( 2 ( c ( c + 1 ) 2 ) 2 + 3 c ( c + 1 ) ( 2 c + 1 ) 6 + c ( c + 1 ) 2 ) [ ∑ k = 1 n k = n ( n + 1 ) 2 ] = 1 12 ∑ c = 1 10 ( c 4 + 2 c 3 + c 2 + 2 c 3 + 3 c 2 + c + c 2 + c ) = 1 12 ∑ c = 1 10 ( c 4 + 4 c 3 + 5 c 2 + 2 c ) [ ∑ k = 1 n k 4 = 6 n 5 + 15 n 4 + 10 n 3 − n 30 ] = 1 12 ( 759990 30 + 4 ( 10 ( 11 ) 2 ) 2 + 5 ( 10 ) ( 11 ) ( 21 ) 6 + 2 ( 10 ) ( 11 ) 2 ) = 1 12 ( 25333 + 12100 + 1925 + 110 ) = 3289 \begin{array} {rll} S & = \displaystyle \sum_{c=1}^{10} \sum_{b=1}^c \sum_{a=1}^b a^2 & \small \displaystyle \color{#3D99F6} {\left[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\right]} \\ & = \displaystyle \sum_{c=1}^{10} \sum_{b=1}^c \frac{b(b+1)(2b+1)}{6} \\ & =\displaystyle \sum_{c=1}^{10} \sum_{b=1}^c \frac{2b^3+3b^2+b}{6} & \small \displaystyle \color{#3D99F6} {\left[ \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 \right]} \\ & =\displaystyle \sum_{c=1}^{10} \frac{1}{6} \left( 2\left(\frac{c(c+1)}{2} \right)^2 + \frac{3c(c+1)(2c+1)}{6} + \frac{c(c+1)}{2} \right) & \small \displaystyle \color{#3D99F6} {\left[ \sum_{k=1}^n k = \frac{n(n+1)}{2} \right]} \\ & = \displaystyle \frac{1}{12} \sum_{c=1}^{10} \left(c^4+2c^3+c^2 + 2c^3+3c^2+c + c^2+c \right) \\ & = \displaystyle \frac{1}{12} \sum_{c=1}^{10} \left(c^4+4c^3+5c^2+2c \right) & \small \displaystyle \color{#3D99F6} {\left[ \sum_{k=1}^n k^4 = \frac{6n^5+15n^4+10n^3-n}{30} \right]} \\ & = \frac{1}{12} \left(\frac{759990}{30} + 4\left(\frac{10(11)}{2}\right)^2 + \frac{5(10)(11)(21)}{6} + \frac{2(10)(11)}{2} \right) \\ & = \displaystyle \frac{1}{12} \left(25333 + 12100+1925+110 \right) = \boxed{3289} \end{array}

22 Helpful 0 Interesting 0 Brilliant 0 Confused

The summation may be simplified a bit using the result ∑ k = 1 n k ( k + 1 ) … ( k + r ) = n ( n + 1 ) … ( n + r ) ( n + r + 1 ) r + 2 \sum_{k=1}^{n}k(k+1)\ldots(k+r)=\frac{n(n+1)\ldots (n+r)(n+r+1)}{r+2} .

Example: b ( b + 1 ) ( 2 b + 1 ) 6 = b ( b + 1 ) ( b + 2 ) 3 − b ( b + 1 ) 2 \frac{b(b+1)(2b+1)}{6}=\frac{b(b+1)(b+2)}{3}-\frac{b(b+1)}{2} .

Chan Lye Lee - 5 years, 8 months ago

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Thanks, you are right.

Chew-Seong Cheong - 5 years, 8 months ago

Sir , what is the x x doing in 5 t h 5^{th} and 6 t h 6^{th} line of your solution ?

Akshat Sharda - 5 years, 8 months ago

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Thanks. Typos

Chew-Seong Cheong - 5 years, 8 months ago

Same solution, but now I'm wondering if there is an elegant Combinatoric solution.

Alan Yan - 5 years, 8 months ago

Sir, is there a way to calculate ∑ x = 1 n x i \sum \limits_{x=1}^n x^{i} for any integral i i

Krutarth Patel - 5 years, 8 months ago

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You can refer to power sum .

Chew-Seong Cheong - 5 years, 8 months ago

That's how I did but in more messy way .. coasted me 1 trial :)

Ahmed Obaiedallah - 5 years, 8 months ago
Nikola Djuric
Dec 14, 2015

=1+(1+(1+4))+(1+(1+4)+(1+4+9))+ …(1+(1+4)+...+(1+4+9+...+100))= 10×1+9×(1+4)+8×(1+4+9)+...+ 1×(1+4+9+ …+100)= 1×(10+9+8+...+1)+4×(9+8+7+...+1)+ …+100×1=1×10×11/2+4×9×10/2+...+ 100×1×2/2=55+180+324+448+525+ 540+490+384+243+100=235+772+ 1065+874+343=1109+1065+1115=3289

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Oct 9, 2015

s:=0;

FOR c:=1 TO 10 DO

FOR b:=1 TO c DO

FOR a:=1 TO b DO 

       s:=s+SQR(a);

Just a simple routine, s = 3289.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

In python: 

1
 
sum(sum(sum(k**2 for k in range(1,m+1)) for m in range(1,n+1)) for n in range(1,11))

Abdelhamid Saadi - 5 years, 8 months ago

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