Reorganizing A Sum

Algebra Level 4

Consider the sum:

S = 1 + 2 ( 1 − x ) + 3 ( 1 − x ) ( 1 − 2 x ) + n ( 1 − x ) ( 1 − 2 x ) . . . ( 1 − ( n − 1 ) x ) S = 1 + 2(1-x) + 3(1-x)(1-2x) + n(1-x)(1-2x)...(1-(n-1)x)

If S S can be written as:

S = 1 x [ 1 − ( 1 − a 1 x ) ( 1 − a 2 x ) ( 1 − a 3 x ) ⋯ ( 1 − a n x ) ] S = \dfrac{1}{x} \left[1-(1-a_1x)(1-a_2x)(1-a_3x)\cdots(1-a_nx) \right]

Find : ∑ i = 1 n a i \displaystyle \sum_{i=1}^n a_i

( n − 1 ) n 2 \dfrac{(n-1)n}{2} n ( n + 1 ) ( 2 n + 1 ) 6 \dfrac{n(n+1)(2n+1)}{6} n ( n + 1 ) 2 \dfrac{n(n+1)}{2} ( n ( n + 1 ) 2 ) 2 \left( \dfrac{n(n+1)}{2} \right)^2

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Priyansh Sangule by Priyansh Sangule , 24, India

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1 solution

In the first expression for S S the constant term will be 1 + 2 + . . . + n = n ( n + 1 ) 2 1 + 2 + ... + n = \frac{n(n + 1)}{2} .

In the second expression for S S , the constant term will be the negative of the coefficient of x x in the expansion of ( 1 − a 1 x ) ( 1 − a 2 x ) . . . . . ( 1 − a n x ) (1 - a_{1}x)(1 - a_{2}x) ..... (1 - a_{n}x) , (due to the 1 x \frac{1}{x} factor at the beginning of the expression). Now by Vieta's rule we know that this will just be ∑ i = 1 n a i \sum_{i=1}^{n} a_{i} .

Thus, upon comparing the two expressions, we can conclude that the correct option is n ( n + 1 ) 2 \boxed{\dfrac{n(n + 1)}{2}} .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Its wrong and I can prove it

Akshay Raj - 6 years, 6 months ago

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then prove it

Dinesh Srikakulapu - 6 years, 6 months ago

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