The formula of general term of mathematical sequence In the sequence {an}, a (1) = 1 / 2, a (n + 1) = a (n) + n ^ 2 Please master quickly help me solve, brackets inside the corner mark

The formula of general term of mathematical sequence In the sequence {an}, a (1) = 1 / 2, a (n + 1) = a (n) + n ^ 2 Please master quickly help me solve, brackets inside the corner mark

a(n)-a(n-1)=(n-1)^2a(n-1)-a(n-2)=(n-2)^2…… A (2) - A (1) = 1a (1) = 1 / 2 plus C (m, n) means to take m individuals from N and combine them. A (n) = 1 / 2 + ∑ (n-1) ^ 2 = 1 / 2 + ∑ [2C (2, n-1) + C (1, n-1)] = 1 / 2 + 2 ∑ C (2, n-1) + ∑ C (1, n-1) = 1 / 2 + 2C (3, n) + C (2, n)
8 (3x-6) = 6 (4x-7) - 3 (2x + 1) (to be detailed)
24x-48=24x-42-6x-3
24x-48=18x-45
6x=3
x=0.5
General term formula of mathematical sequence
Er, A1 = 1, A2 = 3, A3 = 9, A4 = 27, how to write the general term formula? And how to write the general term formula for those non equal difference and equal ratio sequences? Ask for advice! (for example, A2 = 8, A4 = 14, a8 = 26, a16 = 44)
Equal ratio summation: SN = a1 + A2 + a3 +. + an ① when Q ≠ 1, Sn = A1 (1-Q ^ n) / (1-Q) or Sn = (A1 an × q) / (1-Q) ② when q = 1, Sn = n × A1 (q = 1), where Q is the ratio constant, which is 3 in the example
An = A1 × Q & # 710; (n-1): an = 1 × 3 & # 710; (n-1)
Decomposition factor: (x2 + 4x + 8) 2 + 3x (x2 + 4x + 8) + 2x2
Let x2 + 4x + 8 = y, then the original formula = Y2 + 3xy + 2x2 = (y + 2x) (y + x) = (x2 + 6x + 8) (x2 + 5x + 8) = (x + 2) (x + 4) (x2 + 5x + 8)
A general term formula of sequence - 1 / 2,2 / 3, - 3 / 4
an=(-1)^n*n/(n+1)
[(-1)^n]n/(n+1)
General term formula: - N-1 of n
an=(-1)^n*n/(n+1)
Then observe that the numerator is a continuous positive integer starting from 1, and the denominator starts from 2. We can get the n power of (- 1) multiplied by N + 1 / n
8(-2x)^4-(-3x^2)^2-【-(4x^2)^2】+3(-x)^3*x
8(-2x)^4-(-3x^2)^2-【-(4x^2)^2】+3(-x)^3*x
=128x^4-9x^4+16x^4-3x^4
=132x^4
8(-2x)^4-(-3x^2)^2-【-(4x^2)^2】+3(-x)^3*x
=8×2^4×x^4-3^3×x^4+4^2x^4-3x^3×x
=128x^4-9x^4+16x^4-3x^4
=132x^4
One hundred and forty-four
Sequence: 2-178; - 1 / 2,3-178; - 1 / 3,4-178; - 1 / 4,5-178; - 1 / 5 A general term formula of?
(n+1)^2-1/(n+1)
This is observed
solution
General term an = [(n + 1) ² - 1] / (n + 1) = (n + 1) - 1 / (n + 1)
an=（n+1）^2-1/（1+n）
It's not a process
If you want to, guess first and then use mathematical induction
Notice the change of A1 A2 A3 and guess the answer is
(n+1)^2-1/(n+1)
And then use mathematical induction to verify it,
1. When n = 1 is true
2. When n = k, it is assumed that the above formula holds
So when n = K + 1, it holds
High school. If you're a high school student, it's easy
It's not a process
If you want to, guess first and then use mathematical induction
Notice the change of A1 A2 A3 and guess the answer is
(n+1)^2-1/(n+1)
And then use mathematical induction to verify it,
1. When n = 1 is true
2. When n = k, it is assumed that the above formula holds
So when n = K + 1, it holds
What you learned in high school, if you're a high school student, it's easy to put it away
If a line passing through point (1,0) is tangent to the curves y = X3 and y = AX2 + 154x-9, then a is equal to___ .
Let the tangent equation at any point (x0, X03) on the curve y = X3 be y-x03 = 3x02 (x-x0) and (1, 0) be substituted into the equation to get x0 = 0 or x0 = 32. ① when x0 = 0, the tangent equation is y = 0, then AX2 + 154x-9 = 0, △ = (154) 2-4a × (- 9) = 0 {a = - 2564. ② when x0 = 32, the tangent equation is y = 0
A general term formula of sequence 1,3,6,10?
a1=1
a2=a1+2
a3=a2+3
a4=a3+4
.
an=a(n-1)+n
By adding the above formulas, we get
an=1+2+3+...+n=n(n+1)/2
(AX + b) (Cx + D) + (ax-b) & # 178; X power of a = 4, y power of a = 6, find 4x-3y power of a and 4x + 3Y power of A
(AX + b) (Cx + D) + (ax-b) &# 178; is an expression, which is an expression a's x power = 4, a's y power = 6, finding a's 4x-3y power and a's 4x + 3Y power (two are not the same expression)
﹙ax＋b﹚﹙cx＋d﹚＋﹙ax－b﹚²
=acx²+bcx+adx+bd+ax²-2abx+b²
X power of a = 4, y power of a = 6 find 4x-3y power of a and 4x + 3Y power of a
4x-3y power of a
=4X power of a △ 3Y power of a
=(x power of a) ^ 4 ÷ (y power of a) ^ 3
=4^4÷6^3
=256÷216
=32/27
4X + 3Y power of a
=4X power of a * 3Y power of a
=(x power of a) ^ 4 * (y power of a) ^ 3
=4^4*6^3
=256*216
=55296