Use the principle of mathematical induction to prove that

$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + n{\left( {\frac{1}{2}} \right)^{n - 1}} = 4 - \frac{{n + 2}}{{{2^{n - 1}}}}$, where $n \in {\mathbb{Z}^ + }$.

if $n = 1$

${\text{LHS}} = 1\,{\text{;}}\,\,{\text{RHS}} = 4 - \frac{3}{{{2^0}}} = 4 - 3 = 1$     M1

hence true for $n = 1$

assume true for $n = k$     M1

Note: Assumption of truth must be present. Following marks are not dependent on the first two M1 marks.

so $1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} = 4 - \frac{{k + 2}}{{{2^{k - 1}}}}$

if $n = k + 1$

$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}$

$= 4 - \frac{{k + 2}}{{{2^{k - 1}}}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}$      M1A1

finding a common denominator for the two fractions      M1

$= 4 - \frac{{2\left( {k + 2} \right)}}{{{2^k}}} + \frac{{k + 1}}{{{2^k}}}$

$= 4 - \frac{{2\left( {k + 2} \right) - \left( {k + 1} \right)}}{{{2^k}}} = 4 - \frac{{k + 3}}{{{2^k}}}\left( { = 4 - \frac{{\left( {k + 1} \right) + 2}}{{{2^{\left( {k + 1} \right) - 1}}}}} \right)$     A1

hence if true for $n = k$ then also true for $n = k + 1$, as true for $n = 1$, so true (for all $n \in {\mathbb{Z}^ + }$)     R1

Note: Award the final R1 only if the first four marks have been awarded.

[7 marks]