Question

Acest exercitiu din imagine doresc rezolvare!​

Answer 1

Răspuns:

∑(1 k)

(k³ k(k+1))

a11= 1+1+...+1 = n ori

a12=1+2+...+n=n(n+1)/2

a21=1³+2³+...+n³=[n(n+1)]²/4

a22=∑k(k+1)=∑k²+∑k

∑k²=1²+2²+,,,+n²=n(n+1)(2n+1)/6

A=(n n(n+1)/2)

[n(n+1)]²/4 n(n+1)(2n+1)/6)

b)A=∑(1 2^k*3^k+1)

(2^k 2^k*3^(-k))

a11=1+1=...+1=n

∑2^k*3^k+1=3∑6^k=3(6+6²+6^3+...+6^n)=ai o progresie geometrica de ratie 6=suma acestei progresii=

3*(6*(6^n-1)/(6-1)=3*6(6^n-1)/5=18(6^n-1)/5

a21=2^1+2^2+...+2^n=progresie geometrica de ratie 2=

2*(2^n-1)/(2-1)=2*(2^n-1)

2^k*3^-k=(2/3)^k

a22=∑(2/3)^k=(2/3)+(2/3)^2+....+(2/3)6^n=progresie geometrica de ratiie (2/3)=

(2/3*[(2/3)^n-1]/(1-2/3)=2/3*[(2/3)^n-1]/(1/3)=

2[(2/3)^n-1]

construiesti matricea A cu termenii a11 a12,a21 a22 calculati mai sus

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