Given that $\sum\limits_{r=1}^{n}{{{u}_{r}}}=\frac{1}{2}n\left( 3n+1 \right)$, show that ${{u}_{n}}=

Question

Given that $\sum\limits_{r=1}^{n}{{{u}_{r}}}=\frac{1}{2}n\left( 3n+1 \right)$, show that ${{u}_{n}}=3n-1$. Hence find, in terms of $N$, $\sum\limits_{r=1}^{N}{{{2}^{{{u}_{r}}}}}$, $\sum\limits_{r=1}^{N}{{{\left( {{u}_{r}} \right)}^{2}}}$. [You may make use of the result $\sum\limits_{r=1}^{N}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.]

Accepted Answer

$u_n = 3n - 1$; $\frac{4}{7}(2^{3N} - 1)$; $\frac{3}{2}N(N+1)(2N-1)$

2017 TJC PROMO Q11

2017 TJC PROMO Q11

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