Let a1 = b1 = 1, an = a_(n–1 )+ 2 and b_n = a_n + b_(n–1) for every natural number n ≥ 2 . Then

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Let a1 = b1 = 1, an = an–1 + 2 and bn = an + bn–1 for every natural number n ≥ 2. The

let a1=b1=1 and an=an−1+(n−1),bn=bn−1+an−1,∀n≥2.If S=10∑n−1bn2n and T= 8∑n−1n2n−1,then 27(2S−T) is

Let a1 = b1 =1 and an = an-1 + (n-1), bn = bn-1 +an-1 ......... JEE Mains 2023, Jan 29

Let a1=b1=1&an=a(n-1) +(n-1), bn=b(n—1)+a(n-1), if S=sigma n=1 to 10 bn/2^n&T=sigma n=1to8 n/2(n-1),

Summation Formulas You Need to Know #Shorts #math #maths #mathematics

Let an=bn+1/bn, where b1=1, b2=1 and bn+2=bn+1/bn, n∈N. | B Solutions

Prove that a^n - b^n ﹤ (a-b)na^n-1 (given a ﹥ b ﹥ 0 and any integer n ﹥ 1) [ILIEKMATHPHYSICS]

Let A = [ 1 1/51 0 1]. If B= [1 2 -1 -1 ]A[-1 -2 1 1 ] then summation n=1 to n=50 B^n is equal to