Dirty Sum

Algebra Level 5

S = ∑ x = 1 100 x 2 e x + 1 \large S=\sum_{x=1}^{100}x^{2}e^{x+1}

If S = p × 1 0 46 S=p\times10^{46} , find p p rounded to the nearest integer.

Hint : Consider using the telescoping sum technique. Approximating the sum as an integral would not yield the correct result. Using a computer program is not recommended since it would defeat the purpose of this question, but a calculator is necessary.


The answer is 114.

Sam Zhou by Sam Zhou , 19, Canada

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1 solution

Sam Zhou
Aug 7, 2019

Let x 2 e x + 1 = a x + 1 − a x x^{2}e^{x+1}=a_{x+1}-a_{x} , and let a x = e x b x a_{x}=e^{x}b_{x} .

This leads to x 2 e x + 1 = e x + 1 b x + 1 − e x b x = e x + 1 ( b x + 1 − b x e ) x^{2}e^{x+1}=e^{x+1}b_{x+1}-e^{x}b_{x}=e^{x+1}(b_{x+1}-\frac{b_{x}}{e}) . Therefore, b x + 1 − b x e = x 2 b_{x+1}-\frac{b_{x}}{e}=x^{2} .

b x b_{x} must then be quadratic. Set b x b_{x} as A x 2 + B x + C Ax^{2}+Bx+C . We will get A ( x + 1 ) 2 + B ( x + 1 ) + C − A x 2 + B x + C e = x 2 A(x+1)^{2}+B(x+1)+C-\frac{Ax^{2}+Bx+C}{e}=x^{2} .

Rearranging the equation, we get ( A e − 1 e ) x 2 + ( 2 A + B e − 1 e ) x + ( A + B + C e − 1 e ) = x 2 (A\frac{e-1}{e})x^{2}+(2A+B\frac{e-1}{e})x+(A+B+C\frac{e-1}{e})=x^{2} .

Let m = e e − 1 m=\frac{e}{e-1} to simplify the expression. Separating the coefficients, we get:

A m x 2 = x 2 \frac{A}{m}x^{2}=x^{2}

( 2 A + B m ) x = 0 (2A+\frac{B}{m})x=0

A + B + C m = 0 A+B+\frac{C}{m}=0

Solving each equation, we get A = m , B = − 2 m 2 , C = 2 m 3 − m 2 A=m,B=-2m^{2},C=2m^{3}-m^{2} .

Therefore, S = ∑ x = 1 100 a x + 1 − a x = a 101 − a 1 = e 101 ( 10 1 2 m − 2 ( 101 ) m 2 + 2 m 3 − m 2 ) − e ( 1 2 m − 2 ( 1 ) m 2 + 2 m 3 − m 2 ) = 1.14265... × 1 0 48 = 114.265... × 1 0 46 S=\sum_{x=1}^{100}a_{x+1}-a_{x}=a_{101}-a_{1}=e^{101}(101^{2}m-2(101)m^{2}+2m^{3}-m^{2})-e(1^{2}m-2(1)m^{2}+2m^{3}-m^{2})=1.14265...\times10^{48}=114.265...\times10^{46} , giving the answer of 114 \boxed{114} .

0 Helpful 3 Interesting 9 Brilliant 0 Confused

It can be easily but rigorously done by taking the function A ( x ) = e ∑ 0 100 ( x e ) n A(x)=e\sum_{0}^{100} (xe)^n . Then , S = A ′ ′ ( 1 ) + A ′ ( 1 ) S=A''(1)+A'(1) .

Alapan Das - 1 year, 10 months ago

Can someone help me I am not able to do any hard problems like this but can easily do medium level problems.Where can I learn all these theoroms?

Aadit Padhi - 1 year, 10 months ago

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By practicing man....by trying, falling and learning, it may seem poetic...but this is just another average joe speaking

Arnav Das - 1 year, 9 months ago

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