Question

Sum the series sina+5sin2a+7sin3a+...............to n terms

Answer 1

Answer:

3Sina + 5Sin2a + 7Sin3a +..........................to n terms = (Sina + (2n+3)Sin(na) - (2n+1)Sin(n+1)a ) / 4Sin²(a/2)

Step-by-step explanation:

Correct question is

3Sina + 5Sin2a + 7Sin3a +..........................to n terms

n term = (2n+1)Sin(na)

Let say Sum = S

S = 3Sina + 5Sin2a + 7Sin3a +..........................+ (2n+1)Sin(na)

Let say

C = 3Cosa + 5Cos2a + 7Cos3a +..........................+ (2n+1)Cos(na)

using $e^{ia} = Cosa + iSina$

C + iS  = $3e^{ia} + 5e^{i2a} + 7e^{3a} + ............. + (2n+1)e^{ina}$

to simplify let say  x = $e^{ia}$

C + iS =  3x + 5x² + 7x³ + .....................+ (2n+1)xⁿ    - eq 1

Multiplying both sides by x

(C + iS)x = 3x² + 5x³ + 7x⁴ + .....................+ (2n+1)x⁽ⁿ⁺¹⁾    - eq 2

Eq 1 - Eq 2

=> (C + iS)(1-x) = 3x + 2x² + 2x³ +...................+ 2xⁿ  - (2n+1)x⁽ⁿ⁺¹⁾

=> (C + iS)(1-x) = 3x + 2x²(1 + x +...................+ xⁿ⁻²)  - (2n+1)x⁽ⁿ⁺¹⁾

1 + x +...................+ xⁿ⁻² = 1(xⁿ⁻¹ - 1) / (x -1)   (using summation formula of GP)

a =1 , r = x  , terms = n-1

= (1 - xⁿ⁻¹)/(1-x)   ( changing sign of numerator & denominator)

=> (C + iS)(1-x) = 3x + 2x²(1 - xⁿ⁻¹)/(1-x)  - (2n+1)x⁽ⁿ⁺¹⁾

=> (C + iS) = 3x/(1-x)  + 2x²(1 - xⁿ⁻¹)/(1-x)²  - (2n+1)x⁽ⁿ⁺¹⁾/(1-x)

=> C + iS = (3x - x² - (2n+3)xⁿ⁺¹ +(2n+1)xⁿ⁺²)/(1-x)²

(1-x)² =  $(1 - e^{ia})^2 = (e^{\frac{ia}{2}}(e^{\frac{-ia}{2}} - e^{\frac{ia}{2}}))^2 = e^{ia}(e^{\frac{-ia}{2}} - e^{\frac{ia}{2}})^2 = x(e^{\frac{-ia}{2}} - e^{\frac{ia}{2}})^2$

=> C + iS = (3 - x - (2n+3)xⁿ +(2n+1)xⁿ⁺¹)/y

y = $(e^{\frac{-ia}{2}} - e^{\frac{ia}{2}})^2$

y = (cos(-a/2) + iSin(-a/2)  - Cos(a/2) - iSin(a/2) )²

=> y = (Cos(a/2)  - i(sin(a/2) -Cos(a/2) -iSin(a/2))²

=> y = (-2isin(a/2)²

=> y = 4i²Sin²(a/2)

=> y = - 4Sin²(a/2)

=> C + iS = (3 - (cosa + isina) - (2n+3)(cos(na) + iSin(na)) +(2n+1)(Cos(n+1)a + iSin(n+1)a/y

Equating i terms to get S

S =  (-sina -(2n+3)Sin(na) + (2n+1)Sin(n+1)a ) /  - 4Sin²(a/2)

=> S = (Sina + (2n+3)Sin(na) - (2n+1)Sin(n+1)a) / 4Sin²(a/2)