Problem 6
let tn be the nth triangular number, qn be the nth square number, and pn be the nth pentagonal number.
(a) show that q6 + p5 = 3t5 +q5 +1.
(b) show that qn+1 + pn = 3tn +qn + 1 for all positive integers n.

See below.

Step-by-step explanation:

The nth triangular number is Tn = n(n + 1)/2, the nth square number Qn = n^2 and the nth pentagonal number Pn = (3n^2 - n) / 2.

(a.  Q6 + P5 = 6^2 + (3(5)^2 - 5)/2

= 36 + 35 = 71.

3T5 + Q5 + 1 =   3* 5(5 + 1) / 2 + 5^2 + 1

= 90 / 2 + 26

= 71.

So  Q6 + P5 = 3T5 +Q5 +1.

(b) Qn+1 + Pn  = (n + 1)^2 + (3n^2 - n) / 2

=  n^2 + 2n + 1 + (3n^2 - n) / 2

= n^2 + 2n + 1  + 1.5n^2 - 0.5n

= 2.5n^2 + 1.5n + 1.

3Tn +Qn + 1 = 3n(n + 1) / 2  + n^2 + 1

=  3n^2 + 3n / 2 + n^2 + 1

= 1.5n^2 + 1.5n + n^2 + 1

= 2.5n^2 + 1.5n + 1.

Therefore they are equal for all positive n.

circumference divided by 3.141592654 then timesthe radius twice

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice

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step-by-step explanation: