Question

Find the sum of the 'n' terms of the series 1 + (1 + 3) + (1 + 3 + 5) +​

Answer 1

soory i have comed here hor points

Answer 2

Answer:

We have a  type of AP ,

        1 + ( 1 + 3 ) + ( 1 + 3 + 5 ) ----------

      = 1 + 4 + 9 + 16 + ------------

      = 1² + 2² + 3² + 4² + ------------+ n²

       ∑n² = $\frac{n(n+1)(2n+1)}{6}$   (Ans.)

ii)  1³ + 2³ + 3³ + 4³ + ---------- + n³

      ∑n³ = $[\frac{n(n+1)}{2}] ^{2}$  

iii) 1⁴ + 2⁴  + 3⁴ + 4⁴ + ---------- + n⁴

    ∑n⁴ = ∑n² × $(\frac{3n^2 + 3n -1}{5})$  

iv) 1⁵ + 2⁵ + 3⁵ + 4⁵ + ---------- + n⁵

    ∑n⁵ = ∑n³ × $(\frac{2n^2+2n-1}{3} )$

   

by putting the value of n in above formula , you get your answer .

                                            Hope it helps

                                please mark as brainliest