20 uždavinys22 uždavinys

Sprendimas:
S1=4⋅12+4⋅1=4+4=8
Atsakymas: 8

Sprendimas:
S2n=4⋅(2⋅n)2+4⋅(2⋅n)
Sn=4⋅n2+4⋅n
4* ( 2* n)^2+ 4* ( 2* n) =
* ( 4* n^2+ 4* n)

4* ( 2* n)^2+ 4* ( 2* n) = * ( 4* n^2+ 4* n)4⋅(2⋅n)2+4⋅(2⋅n) = 311⋅(4⋅n2+4⋅n) 
16* n^2+ 4* ( 2* n) = * ( 4* n^2+ 4* n)16⋅n2+4⋅(2⋅n) = 311⋅(4⋅n2+4⋅n) 
16* n^2+ 8* n = * ( 4* n^2+ 4* n)16⋅n2+8⋅n = 311⋅(4⋅n2+4⋅n) 
16* n^2+ 8* n = * 4* n^2+ * 4* n16⋅n2+8⋅n = 311⋅4⋅n2+311⋅4⋅n 
16* n^2+ 8* n = + * 4* n16⋅n2+8⋅n = 344⋅n2+311⋅4⋅n 
16* n^2+ 8* n = + 16⋅n2+8⋅n = 344⋅n2+344⋅n 
16* n^2- + 8* n = 16⋅n2−344⋅n2+8⋅n = 344⋅n 
16* n^2- + 8* n- = 016⋅n2−344⋅n2+8⋅n−344⋅n = 0 
+ 8* n- = 034⋅n2+8⋅n−344⋅n = 0 
- = 034⋅n2−320⋅n = 0 
- = 034⋅n−320 = 0 
= 0+ 34⋅n = 0+320 
= 34⋅n = 320 
4* n = 4⋅n = 320⋅3 
4* n = 204⋅n = 20 
n = n = 420 
Atsakymas: n = 5
20 uždavinys22 uždavinys