JEE Main 202329 Jan, Evening shiftMathematicsMediumSequence and Series179× since 2002Let a1=b1=1a_1=b_1=1a1=b1=1 and an=an−1+(n−1),bn=bn−1+an−1,∀n≥2{a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2an=an−1+(n−1),bn=bn−1+an−1,∀n≥2. If S=∑n=110bn2nS = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}}S=n=1∑102nbn and T=∑n=18n2n−1T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}}T=n=1∑82n−1n, then 27(2S−T){2^7}(2S - T)27(2S−T) is equal to ____________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
