Math, asked by jas2331, 8 months ago

Sum to n terms the series
3 sin a + 5 sin 2a + 7 sin 3a +.......

Answers

Answered by Anonymous

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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

C + iS =

to simplify let say x =

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 =

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

y =

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)