An algebra problem by Anshuman Shreshth

Algebra Level 3

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The answer is 9.

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3 solutions

Prasun Biswas
Feb 4, 2015

What is given to us is:

i = 1 K i 3 i = 1 K i = 1980 \sum_{i=1}^K i^3 - \sum_{i=1}^K i = 1980

We will use two identities: i = 1 n i 3 = ( n ( n + 1 ) 2 ) 2 and i = 1 n i = n ( n + 1 ) 2 \displaystyle \sum_{i=1}^n i^3 = \left(\dfrac{n(n+1)}{2}\right)^2\quad \textrm{and}\quad \displaystyle \sum_{i=1}^n i = \dfrac{n(n+1)}{2}

Let X = K ( K + 1 ) 2 X=\dfrac{K(K+1)}{2} . The given sum can now be written as:

X 2 X 1980 = 0 ( X 45 ) ( X + 44 ) = 1980 X = 45 X^2-X-1980=0 \implies (X-45)(X+44)=1980 \implies X=45

The negative solution is rejected since K K cannot be negative and hence X X also cannot be negative. Similarly, we'll reject the negative K K in the next step. Now,

X = K ( K + 1 ) 2 = 45 K 2 + K 90 = 0 ( K 9 ) ( K + 10 ) = 0 K = 9 X=\dfrac{K(K+1)}{2}=45\\ \implies K^2+K-90=0 \implies (K-9)(K+10)=0 \implies \boxed{K=9}

Out of curiosity, what would be the fastest way to find the required factors to factorise X 2 X 1980 = 0 X^{2}-X-1980=0 ? I listed all the possible factors, and the 1 7 th 17^{\text{th}} pair of factors came to be the required ones. I spent about five minutes listing all the factors.

Omkar Kulkarni - 6 years, 2 months ago

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@Omkar Kulkarni , it's a quadratic equation. Just use the quadratic formula and you'll get that the roots are x = 45 , ( 44 ) x=45,(-44) .

x = 1 ± 1 + 7920 2 = 1 ± 89 2 = 45 , ( 44 ) x=\frac{1\pm\sqrt{1+7920}}{2}=\frac{1\pm 89}{2}=45,(-44)

That's the fastest one can go with a calculator. There's also the method of middle term factorization that can be used.

Prasun Biswas - 6 years, 1 month ago
William Isoroku
Feb 10, 2015

Use the formula: 1 3 + 2 3 + 3 3 + . . . . . + n 3 = ( n ( n + 1 ) 2 ) 2 { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ n }^{ 3 }={ (\frac { n(n+1) }{ 2 } ) }^{ 2 } and + 2 + 3 + . . . . . + n = n ( n + 1 ) 2 +2+3+.....+n=\frac { n(n+1) }{ 2 }

As the problem follows: ( n ( n + 1 ) 2 ) 2 n ( n + 1 ) 2 = 1980 { (\frac { n(n+1) }{ 2 } ) }^{ 2 }-\frac { n(n+1) }{ 2 } =1980

Combining the fraction and simplifying: n 4 + 2 n 3 n 2 n 7920 = 0 { n }^{ 4 }+2{ n }^{ 3 }-{ n }^{ 2 }-n-7920=0

Graph it and the approximate positive root is 9.997 9.997 which is rounded to 10 \boxed{10}

Saket Sumant
Apr 2, 2015

anshuman shreshth, the conclusion which you reached in the class is already a formulae.
that is the sum of cube of terms of an a.p. is equal to square of the sum of the terms of the a.p. .

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