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Question: f(n), g(n), and h(n) are second degree polynomials in n having n + 2 as a common factor then value o...

f(n), g(n), and h(n) are second degree polynomials in n having n + 2 as a common factor then value of ∑r=1nΔr\sum_{r = 1}^{n}\Delta_{r}is, where

∆r= ∣2r+16n(n+2)f(n)2r−13.2n+1−6g(n)r(n−r+1)n(n+1)(n+2)h(n)∣\left| \begin{matrix} 2r + 1 & 6n(n + 2) & f(n) \\ 2^{r - 1} & 3.2^{n + 1} - 6 & g(n) \\ r(n - r + 1) & n(n + 1)(n + 2) & h(n) \end{matrix} \right|

A

2n(n)(n+1)(n+2)6\frac{2^{n}(n)(n + 1)(n + 2)}{6}

B

n(n+1)(n+2)12\frac{n(n + 1)(n + 2)}{12}

C

(2n−1)(n+1)3\frac{(2^{n} - 1)(n + 1)}{3}

D

None of these

Answer

None of these

Explanation

Solution

We have to find the sum of n determinants whose second and third columns are the same for all determinants. Hence, we use the sum of determinants.

∑r=1nΔr\sum_{r = 1}^{n}\Delta_{r}= ∆1 + ∆2 + … + ∆n

= +∣2.1+16n(n+2)f(n)203.2n+1−6g(n)1.(n+1−1)n(n+1)(n+2)h(n)∣\left| \begin{matrix} 2.1 + 1 & 6n(n + 2) & f(n) \\ 2^{0} & 3.2^{n + 1} - 6 & g(n) \\ 1.(n + 1 - 1) & n(n + 1)(n + 2) & h(n) \end{matrix} \right|

+∣2.2+16n(n+2)f(n)213.2n+1−6g(n)2(n+1−2)n(n+1)(n+2)h(n)∣\left| \begin{matrix} 2.2 + 1 & 6n(n + 2) & f(n) \\ 2^{1} & 3.2^{n + 1} - 6 & g(n) \\ 2(n + 1 - 2) & n(n + 1)(n + 2) & h(n) \end{matrix} \right| …to n determinants

=∣2(1+2+3+...+n)+(1+1+...+1)6n(n+2)f(n)20+21+22+...+2n−13.2n+1−6g(n)1.(n+1−1)+2(n+1−2)+...+n(n+1−n)n(n+1)(n+2)h(n)∣\left| \begin{matrix} 2(1 + 2 + 3 + ... + n) + (1 + 1 + ... + 1) & 6n(n + 2) & f(n) \\ 2^{0} + 2^{1} + 2^{2} + ... + 2^{n - 1} & 3.2^{n + 1} - 6 & g(n) \\ 1.(n + 1 - 1) + 2(n + 1 - 2) + ... + n(n + 1 - n) & n(n + 1)(n + 2) & h(n) \end{matrix} \right|

+ …. to n determinants

=∣2.n(n+1)2+n6n(n+2)f(n)2n−12−13(2n+1−2)g(n)(n+1)(1+2+...+n)−(12+22+...+n2)n(n+1)(n+2)h(n)∣\left| \begin{matrix} 2.\frac{n(n + 1)}{2} + n & 6n(n + 2) & f(n) \\ \frac{2^{n} - 1}{2 - 1} & 3(2^{n + 1} - 2) & g(n) \\ (n + 1)(1 + 2 + ... + n) - (1^{2} + 2^{2} + ... + n^{2}) & n(n + 1)(n + 2) & h(n) \end{matrix} \right|

= ∣n(n+2)6n(n+2)f(n)2n−16(2n−1)g(n)(n+1)n(n+1)2−n(n+1)(2n+1)66.n(n+1)(n+2)6h(n)∣\left| \begin{matrix} n(n + 2) & 6n(n + 2) & f(n) \\ 2^{n} - 1 & 6(2^{n} - 1) & g(n) \\ (n + 1)\frac{n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6} & 6.\frac{n(n + 1)(n + 2)}{6} & h(n) \end{matrix} \right|

=6. ∣n(n+2)n(n+2)f(n)2n−12n−1g(n)n(n+1)6(n+2)n(n+1)(n+2)6h(n)∣\left| \begin{matrix} n(n + 2) & n(n + 2) & f(n) \\ 2^{n} - 1 & 2^{n} - 1 & g(n) \\ \frac{n(n + 1)}{6}(n + 2) & \frac{n(n + 1)(n + 2)}{6} & h(n) \end{matrix} \right|

= 6 × 0 (Q C1 ≡ C2) = 0

∴∑r=1nΔr\sum_{r = 1}^{n}\Delta_{r}is independent of n.