Porisma

Worked solution

Recover terms and check the exceptional first term

For each series, find an expression for its nth term unu_n and determine whether the terms form an AP or GP: (a) Sn=2n−1S_n=2^n-1; (b) Sn=n2+3nS_n=n^2+3n; (c) Sn=n2+3n+1S_n=n^2+3n+1, where SnS_n is the sum of the first nn terms.

CJC · Sequences and Series · Example 12

(a) Exponential partial sums

Answer

un=2n−1u_n=2^{n-1} for all n∈Z+n\in\mathbb Z^+; GP with ratio 22.

Subtract consecutive partial sums for n at least two

un=Sn−Sn−1=(2n−1)−(2n−1−1)=2n−2n−1=2n−1(2−1)=2n−1(n≥2).\begin{aligned}u_n&=S_n-S_{n-1}\\&=(2^n-1)-(2^{n-1}-1)\\&=2^n-2^{n-1}\\&=2^{n-1}(2-1)\\&=2^{n-1}\qquad(n\ge2).\end{aligned}

Retain the full source factorisation.

Check the first term separately

u1=S1=21−1=1,21−1=1u_1=S_1=2^1-1=1,\qquad2^{1-1}=1

S1 is the first term. The same formula gives 1 at n=1, so it extends to all positive indices.

Prove the common ratio

un+1un=2n2n−1=2\frac{u_{n+1}}{u_n}=\frac{2^n}{2^{n-1}}=2

The source explicitly checks general neighbouring terms.

How did this part go?

(b) Quadratic partial sums

Answer

un=2n+2u_n=2n+2 for all n∈Z+n\in\mathbb Z^+; AP with difference 22.

Subtract and simplify

un=Sn−Sn−1=(n2+3n)−[(n−1)2+3(n−1)]=(n2+3n)−(n2+n−2)=2n+2(n≥2).\begin{aligned}u_n&=S_n-S_{n-1}\\&=(n^2+3n)-[(n-1)^2+3(n-1)]\\&=(n^2+3n)-(n^2+n-2)\\&=2n+2\qquad(n\ge2).\end{aligned}

The shifted quadratic must be expanded before cancelling.

Check that the formula includes the first term

u1=S1=12+3(1)=4,2(1)+2=4u_1=S_1=1^2+3(1)=4,\qquad2(1)+2=4

The Teacher supplies this first-term check.

Prove the common difference

un+1−un=[2(n+1)+2]−(2n+2)=2n+4−2n−2=2.\begin{aligned}u_{n+1}-u_n&=[2(n+1)+2]-(2n+2)\\&=2n+4-2n-2\\&=2.\end{aligned}

The illustrated source callout asks for a general consecutive-difference or ratio check, rather than merely rewriting into a familiar AP/GP form.

How did this part go?

(c) A constant added to the partial sum

Answer

u1=5u_1=5 and un=2n+2u_n=2n+2 for n≥2n\ge2; neither AP nor GP.

Subtract only where the relation applies

un=Sn−Sn−1=(n2+3n+1)−[(n−1)2+3(n−1)+1]=(n2+3n+1)−[(n2−2n+1)+3n−3+1]=(n2+3n+1)−(n2+n−1)=2n+2(n≥2).\begin{aligned}u_n&=S_n-S_{n-1}\\&=(n^2+3n+1)-[(n-1)^2+3(n-1)+1]\\&=(n^2+3n+1)-[(n^2-2n+1)+3n-3+1]\\&=(n^2+3n+1)-(n^2+n-1)\\&=2n+2\qquad(n\ge2).\end{aligned}

The constant cancels between consecutive partial sums, but this calculation applies from n=2.

Keep the exceptional first term

u1=S1=12+3(1)+1=5≠2(1)+2=4,un={5n=1,2n+2n≥2.u_1=S_1=1^2+3(1)+1=5\ne2(1)+2=4,\qquad u_n=\begin{cases}5&n=1,\\2n+2&n\ge2.\end{cases}

The source check fails at n=1, so retain the piecewise result.

Disprove both classifications

(u1,u2,u3)=(5,6,8),u2−u1=1≠2=u3−u2,u2u1=65≠43=u3u2.\begin{aligned}(u_1,u_2,u_3)&=(5,6,8),\\u_2-u_1=1&\ne2=u_3-u_2,\\\frac{u_2}{u_1}=\frac65&\ne\frac43=\frac{u_3}{u_2}.\end{aligned}

The source compares the first two differences and ratios. These explicit numerical values are authored checks of its inequalities.

How did this part go?

Common mistakes

Subtracting partial sums at n=1 would introduce an unspecified S0. A constant in Sn can change the first term even when it cancels in later differences.