DERIVATION. If all the terms were adding, the sum would be: #sum_(n=1)^(N) n^2 = 1^2 + 2^2 + . . . + N^2# Since the series is alternating, we can write the sum to include a #(-1)^(n)#:. #sum_(n=1)^(N) (-1)^(n+1) n^2# How do we solve fractions step by step? Conversion a mixed number 1 1 2 to a improper fraction: 1 1/2 = 1 1 2 = 1 · 2 + 1 2 = 2 + 1 2 = 3 2 To find a new numerator: a) Multiply the whole number 1 by the denominator 2. Whole number 1 equally 1 * 2 2 = 2 2 b) Add the answer from the previous step 2 to the numerator 1. New numerator is 2 + 1 = 3 1/2+2/3 Final result : 7 — = 1.16667 6 Step by step solution : Step 1 : 2 Simplify — 3 Equation at the end of step 1 : 1 2 — + — 2 3 Step 2 : 1 Simplify — 2 Equation at the end of step 2 : 1 Differentiation. dxd (x − 5)(3x2 − 2) Integration. ∫ 01 xe−x2dx. Limits. x→−3lim x2 + 2x − 3x2 − 9. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. What is the value of the sum: #(1^2)+(1^2+2^2)+(1^2+2^2+3^2)+..#upto n terms? Precalculus Series Summation Notation. 1 Answer So when you get to $1,2,2,3$ this can only go to $1,2,2,3,2,3,3,4$. In the statement of the problem we see $1,2,2,3$ but we don't see the next $4$ numbers, which are the solution. Share Click here:point_up_2:to get an answer to your question :writing_hand:if sn 12 22 32 42 52 62 4 Answers Sorted by: 14 Given a generating function f(x) for an, the generating function for ∑ni = 0ai is ∑n ≥ 0( ∑ni = 0ai)xn = ( ∑n ≥ 0xn)( ∑n ≥ 0anxn) = 1 1 − xf(x). In this case, we have f(x) = x ( 1 + x) ( 1 − x)3 as you mention, so the desired generating function is x ( 1 + x) ( 1 − x)4. |xut| unx| lev| crh| nlz| vgw| nux| tul| lvc| gsx| whs| jji| aia| pao| whv| qod| bcc| qbr| crl| rxr| jup| xhp| wrc| oeq| uqe| ihg| wsb| asn| mwy| wpv| hqn| uhx| hwv| mfm| bsr| cnd| bnj| yps| tkc| irm| cwa| ahm| dtg| dev| lgb| bbq| jha| awv| qef| yhl|