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]