Porisma

Worked solution

Generate terms from a preceding-term recurrence

Given un=2un−1+3u_n=2u_{n-1}+3, u1=2u_1=2, find u2,u3,u4u_2,u_3,u_4 and u20u_{20}.

CJC · Sequences and Series · Example 3

Answer

u2=7u_2=7, u3=17u_3=17, u4=37u_4=37, u20=2621437u_{20}=2621437.

Compare your reasoning

Advance one term at a time

u2=2(2)+3=7,u3=2(7)+3=17,u4=2(17)+3=37.\begin{aligned}u_2&=2(2)+3=7,\\u_3&=2(7)+3=17,\\u_4&=2(17)+3=37.\end{aligned}

Starting with u₁, apply the preceding-term relation from n=2 onwards; each new term uses the preceding computed term.

Follow the source calculator method

On the device shown: (1) MODE → SEQ; (2) y=, set nMin=1; (3) 2nd 7 selects u, enter 2u(n−1)+3 using X,T,θ,n for n; (4) set u(1)=2 and leave u(2) blank; (5) 2nd graph, then Up/Down to inspect terms. The printed table starts 2,7,17,37,77,157,317,637,1277,2557,5117. Continue to index 20 for the requested term.

Check the twentieth term independently

un+3=2(un−1+3),u1+3=5,un=5⋅2n−1−3,u20=5⋅219−3=2621437.\begin{aligned}u_n+3&=2(u_{n-1}+3),\qquad u_1+3=5,\\u_n&=5\cdot2^{n-1}-3,\\u_{20}&=5\cdot2^{19}-3=2621437.\end{aligned}

This shifted-form algebra supplements the source calculator workflow and its printed answer.

How did this question go?

Common mistakes

The first term is u₁. Reaching u₂₀ takes nineteen updates, not twenty.