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Scilla [17]
2 years ago
8

Please help to find an explicit formula for calculating the sum Mn

Mathematics
1 answer:
defon2 years ago
8 0

The explicit formula for calculating the sum is

S_N=\frac{n(n-1)}{2} \cdot\frac{n(n+1)}{2}

The sum of the nth term of a sequence is expressed as;

S_n=\frac{n}{2}(2a+(n-1)d)

a is the first term

d is the common difference

n is the number of terms

For the sequence  0 + 1 + 2 + 3  +...

S_n=\frac{n}{2}(2(0)+(n-1)1)\\S_n= \frac{n}{2}(n-1)\\S_n= \frac{n(n-1)}{2}

Similarly for the sequence:

1 + 2+ 3 + 4+...

S_n=\frac{n}{2}(2(1)+(n-1)1)\\S_n= \frac{n}{2}(2+n-1)\\S_n= \frac{n(n+1)}{2}

Taking the product of the sum to get the explicit formula for calculating the sum

S_N=\frac{n(n-1)}{2} \cdot\frac{n(n+1)}{2}

Learn more here: brainly.com/question/24547297

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Given that (ax^2 + bx + 3) (x + d) = x^3 + 6x^2 + 11x + 12<br> a + 2b - d = ?
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Answer:

Let's solve for a.

(ax2+bx+3)(x+d)=x3+6x2+11x+12a+2b−d

Step 1: Add -12a to both sides.

adx2+ax3+bdx+bx2+3d+3x+−12a=x3+6x2+12a+2b−d+11x+−12a

adx2+ax3+bdx+bx2−12a+3d+3x=x3+6x2+2b−d+11x

Step 2: Add -bdx to both sides.

adx2+ax3+bdx+bx2−12a+3d+3x+−bdx=x3+6x2+2b−d+11x+−bdx

adx2+ax3+bx2−12a+3d+3x=−bdx+x3+6x2+2b−d+11x

Step 3: Add -bx^2 to both sides.

adx2+ax3+bx2−12a+3d+3x+−bx2=−bdx+x3+6x2+2b−d+11x+−bx2

adx2+ax3−12a+3d+3x=−bdx−bx2+x3+6x2+2b−d+11x

Step 4: Add -3d to both sides.

adx2+ax3−12a+3d+3x+−3d=−bdx−bx2+x3+6x2+2b−d+11x+−3d

adx2+ax3−12a+3x=−bdx−bx2+x3+6x2+2b−4d+11x

Step 5: Add -3x to both sides.

adx2+ax3−12a+3x+−3x=−bdx−bx2+x3+6x2+2b−4d+11x+−3x

adx2+ax3−12a=−bdx−bx2+x3+6x2+2b−4d+8x

Step 6: Factor out variable a.

a(dx2+x3−12)=−bdx−bx2+x3+6x2+2b−4d+8x

Step 7: Divide both sides by dx^2+x^3-12.

a(dx2+x3−12)

dx2+x3−12

=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

a=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

Answer:

a=

−bdx−bx2+x3+6x2+2b−4d+8x/

dx2+x3−12

Step-by-step explanation:

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