Show that $d = {\text{lo}}{{\text{g}}_c}\left( q \right)$.

Let $p = {c^2}$ and $q = {c^3}$. Find the value of $\sum\limits_{n = 1}^{20} {{u_n}}$.

valid approach involving addition or subtraction       M1
eg  ${u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} - {u_2}$

correct application of log law      A1
eg  ${\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)$

$d = {\text{lo}}{{\text{g}}_c}\,q$    AG N0

[2 marks]

METHOD 1 (finding ${u_1}$ and d)

recognizing $\sum { = {S_{20}}}$ (seen anywhere)      (A1)

attempt to find ${u_1}$ or d using ${\text{lo}}{{\text{g}}_c}\,{c^k} = k$     (M1)
eg  ${\text{lo}}{{\text{g}}_c}\,c$, ${\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c$, correct value of ${u_1}$ or d

${u_1}$ = 2, d = 3 (seen anywhere)      (A1)(A1)

correct working     (A1)
eg  ${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$

$\sum\limits_{n = 1}^{20} {{u_n}}$ = 610     A1 N2

METHOD 2 (expressing S in terms of c)

recognizing $\sum { = {S_{20}}}$ (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  $10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$

${\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3$  (seen anywhere)     (A1)(A1)

correct working      (A1)

eg  ${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$

$\sum\limits_{n = 1}^{20} {{u_n}}$ = 610     A1 N2

METHOD 3 (expressing S in terms of c)

recognizing $\sum { = {S_{20}}}$ (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  $10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$

correct application of log law     (A1)
eg  $2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)$

correct application of definition of log      (A1)
eg  ${\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57$

correct working     (A1)
eg  ${S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)$

$\sum\limits_{n = 1}^{20} {{u_n}}$ = 610     A1 N2

[6 marks]