 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 4

The special <binomial theorem>
$$
(1+t)^n=\sum_{k=0}^n\binom nk t^k
$$
follows by <mathematical induction>. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).

Termwise integration from $0$ to $1$ gives
$$
\boxed{
\sum_{k=0}^n\frac1{k+1}\binom nk
=\int_0^1(1+t)^n\,dt
=\frac{2^{n+1}-1}{n+1}}.
$$
Replacing $t$ by $-t$ gives
$$
\boxed{
\sum_{k=0}^n\frac{(-1)^k}{k+1}\binom nk
=\int_0^1(1-t)^n\,dt
=\frac1{n+1}}.
$$

Finally,
$$
\sum_{k=1}^n\frac{(-1)^{k+1}}k\binom nk
=\int_0^1\frac{1-(1-x)^n}{x}\,dx.
$$
Writing $y=1-x$ and using the finite <geometric series>,
$$
\frac{1-y^n}{1-y}=1+y+\cdots+y^{n-1},
$$
turns this into
$$
\boxed{\sum_{j=0}^{n-1}\int_0^1y^j\,dy
=1+\frac12+\cdots+\frac1n}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 6D / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 6D / b / Solution

 Ancestors (11)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 6D / b / Solution

 Ancestors (11)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)