Find the value of 1×4+2×5+3×6.n×（n+3）​

n(n+1)(n+5)/3

Step-by-step explanation:

there is no value, as we don't know n.

but we can create a summary formula/ function definition :

this is the sum for k = 1 to n of k×(k+3)

k×(k+3) = k² + 3k

so, the overall sum splits into the sum of k² for k=1 to n, and the sum of 3k for k=1 to n.

and the sum of 3k is 3 times the sum of k for k=1 to n.

Σk² for k=1 to n = [n(n+1)(2n+1)]/6

Σk for k=1 to n = n(n+1)/2

3×Σk for k=1 to n = 3×n(n+1)/2

so, we have a function formula

n(n+1)(2n+1)/6 + 3n(n+1)/2 = n(n+1)(2n+1)/6 + 9n(n+1)/6 =

= n(n+1)(2n+1+9)/6 = n(n+1)(2n+10)/6 = n(n+1)(n+5)/3

you would use (d.) the law of cosines or a^2 = b^2 + c^2 – 2bc(cos a)

, b^2 = a^2 + c^2 – 2ac(cos b), or c^2 = a^2 + b^2 - 2ab(cos c). you wouldn't use b. the law of sines as it is mostly effective when 1side and 2angles are given. a and c aren't the answer for they are the improper versions of the law of cosines. hope this is .