Question

let n - 6+66+666+...+666...66,where there are hundred 6'sin the last term in the sum.how many times does the digit 7 occur in the number n

Answer 1

Sn Sn = 6 / 9 [ 9 + 99 + 999 + ⋯ ]

= 6/ 9 [ ( 10 – 1 ) + ( 102 – 1 ) + ( 103 – 1 ) +⋯ ]

= 6 / 9 [ ( 10 + 102 + 103 + ⋯ ) − ( 1 + 6 9 [ 10 ( 10 n – 1 ) /9 – n ] 1 + 1 + ⋯ ) ]

Answer 2

Thank you for asking this question. Here is your answer:

6+66+666+ (100 terms )

=6(1+11+111+ (100terms)

=(6/9)(9+99+999+(100 terms)

=(2/3){(10–1) + (100–1) + (1000–1) (100 terms}

=(2/3){(10+100+1000+….100terms)-(1+1+(100 terms)

=(2/3){10(10^100 - 1)/9 - 100} (since 10+100 + it is a G.P. So we can use sum to n terms formula)

=(20/27)(10^100 - 1) -200/3

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