Question

please answer this ​

Answer 1

To Prove :-

• CI = P[(1 + R/100)^n - 1]

Proof :-

Let,

→ Principal = P

→ Rate = R% per annum

→ T = 1 year

So,

→ Simple interest for first year SI(1) = (P * R * T)/100 = (P * R * 1)/100 = (PR/100)

then,

→ Amount for one year (A1) = P + SI(1) = P + (PR/100) = P[1 + (R/100)]

now,

→ Principal for second year (P2) = Amount after one (A1) = P[1 + (R/100)]

So,

→ Simple interest for second year SI(2) = (P2 * R * 1)/100 = (P2 * R)/100

then,

→ Amount for second year (A2) = P2 + SI(2) = P2 + (P2 * R)/100 = P2[1 + (R/100)] = P[1 + (R/100)] * [1 + (R/100)] = P[1 + (R/100)]²

similarly, we can conclude that,

→ Amount for n years = P[1 + (R/100)]^n

therefore,

→ CI for n years = Amount - Principal = P[1 + (R/100)]^n - P = P[(1 + R/100)^n - 1]

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