Whole qn? Or part

Whole question except for b(i)

∑ (r = N + 1, to 2N) (2^r + 3^N)

Realise that ∑ 2^r is actually the sum of a GP with first term 2^(N+1) and last term 2^(2N), common ratio 2.

3^N is a constant so just multiply by number of terms.

Number of terms from N+1 to 2N
= N

(Or 2N - (N+1) + 1 = 2N - N - 1 + 1 = N)

Eg. If N is 20, you would be wanting the 21st term to 40th term. Number of terms is 20 or 40 - 21 + 1 = 20. )



So,


∑ (r = N + 1, to 2N) (2^r + 3^N)


= 2^(N+1) (1 - 2^N) / (1 - 2) + N(3^N)

= 2^(N+1) (2^N - 1) + (3^N) N

= 2^(2N+1) - 2^(N+1) + (3^N)N

∑ (r =1, to n terms) ( 2^(-r) (2-3r) )

= ∑ ( f(r) - f(r - 2) )

= ∑ ( (r+2)2^(-r) - (r)2^(-(r-2)) )

= ∑ ( (r+2)/2^r - r/2^(r-2) )


= (1 + 2)/2 - 1/2¹-²

+ (2 + 2)/2² - 2/2²-²

+ (3 + 2)/2³ - 3/2³-²

+ (4 + 2)/2⁴ - 4/2⁴-²

+ (5 + 2)/2^5 - 5/2^(5 - 2)

+...

+ (n + 1)/2ⁿ-¹ - (n - 1)/2ⁿ-¹-²

+ (n + 2)/2ⁿ - n/2ⁿ-²


= 3/2 - 2

+ 1 - 2

+ ⅝ - 3/2

+ 6/16 - 1

+ 7/32 - ⅝

+ ...

+ (n + 1)/2ⁿ-¹ - (n - 1)/2ⁿ-³

+ (n + 2)/2ⁿ - n/2ⁿ-²

(Cancel common terms. Only the 2nd, 4th, 2nd last and 4th last terms are not cancelled)

= -2 - 2 + (n + 1)/2ⁿ-¹ + (n + 2)/2ⁿ

= 2(n + 1)/2ⁿ + (n + 2)/2ⁿ - 4

= (2n + 2 + n + 2)2-ⁿ - 4

= 2-ⁿ(3n + 4) - 4

(shown)

(Edited)

Edited