Solveeit Logo
Login

Question

Question: If \(\sum_{n = 1}^{n}\alpha_{n}\)= an<sup>2</sup> + bn, where a, b are constants and a<sub>1</sub>, ...

If ∑n=1nαn\sum_{n = 1}^{n}\alpha_{n}= an2 + bn, where a, b are constants and a1, a2, a3 Ī {1, 2, 3, ……., 9} and 25a1, 37a2, 49a3 be three digit numbers, then ∣α1α2α357925α137α249α3∣\left| \begin{matrix} \alpha_{1} & \alpha_{2} & \alpha_{3} \\ 5 & 7 & 9 \\ 25\alpha_{1} & 37\alpha_{2} & 49\alpha_{3} \end{matrix} \right|=

A

a1 + a2 + a3

B

a1 – a2 + a3

C

7

D

0

Answer

0

Explanation

Solution

∑n=1n\overset{\underset{\mathbf{n}}{\mathbf{n = 1}}}{\mathbf{\sum}}an = an2 + bn

as sum of a series is a quadratic function of n so this series must be an A.P. so 2a2 = a1 + a3 Applying

R3 ® R3 – (10R2 + R1) ∣α1α2α3579200300400∣\left| \begin{matrix} \alpha_{1} & \alpha_{2} & \alpha_{3} \\ 5 & 7 & 9 \\ 200 & 300 & 400 \end{matrix} \right|

C2 ® C2 – (C1+C32)\left( \frac{C_{1} + C_{3}}{2} \right) = ∣α10α35092000400∣\left| \begin{matrix} \alpha_{1} & 0 & \alpha_{3} \\ 5 & 0 & 9 \\ 200 & 0 & 400 \end{matrix} \right|= 0030