Use the principle of mathematical induction to prove that

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

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

if $n=k+1$

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

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