sum

2 Answers

tn = 2tn-1 + n

now u derive

tn+1 = Sn + (n+1)(n+2)/2

tn+1 = 2ⁿ⁺¹ + 2ⁿ.2 + 2^n-1.3 .........+ n+1

simpler method :

tn can be found usin AGP and u can apply sum of GP for that exp also

will be very long but simple process

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